Trigonometry
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These are the solutions to the WASSCE past questions on the topics in Trigonometry.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided
for some questions.
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Please NOTE: For applicable questions (questions that require the angle in degrees), if you intend to
use the TI-84 Family:
Please make sure you set the MODE to DEGREE.
It is RADIAN by default. So, it needs to be set to DEGREE.
Coterminal Angles
Coterminal angles are angles with the same initial side and the same terminal side.
The concept of coterminal angles helps us to find the equivalent positive angle of a negative angle.
Because an angle in standard position is measured counterclockwise, adding 360° to it accounts for a full
revolution, keeping the direction intact.
Hence for any angle say θ, the coterminal angles are θ + 360k where k is any integer.
Trigonometric Functions
Right Triangle Trigonometry for any angle, say $\theta$
Pneumonic for it is: SOHCAHTOA and cosecHOsecHAcotanAO
$
(1.)\;\; \sin \theta = \dfrac{opp}{hyp} \\[7ex]
(2.)\;\; \cos \theta = \dfrac{adj}{hyp} \\[7ex]
(3.)\;\; \tan \theta = \dfrac{\sin \theta}{\cos \theta} \\[5ex]
= \sin \theta \div \cos \theta \\[3ex]
= \dfrac{opp}{hyp} \div \dfrac{adj}{hyp} \\[5ex]
= \dfrac{opp}{hyp} * \dfrac{hyp}{adj} \\[5ex]
\therefore \tan \theta = \dfrac{opp}{adj} \\[7ex]
(4.)\;\; \csc \theta = \dfrac{1}{\sin \theta} \\[5ex]
= 1 \div \sin \theta \\[3ex]
= 1 \div \dfrac{opp}{hyp} \\[5ex]
\therefore \csc \theta = \dfrac{hyp}{opp} \\[7ex]
(5.)\;\; \sec \theta = \dfrac{1}{\cos \theta} \\[5ex]
= 1 \div \cos \theta \\[3ex]
= 1 \div \dfrac{adj}{hyp} \\[5ex]
\therefore \sec \theta = \dfrac{hyp}{adj} \\[7ex]
(6.)\;\; \cot \theta = \dfrac{1}{\tan \theta} = \dfrac{\cos \theta}{\sin \theta} \\[5ex]
= \cos \theta \div \sin \theta \\[3ex]
= \dfrac{adj}{hyp} \div \dfrac{opp}{hyp} \\[5ex]
= \dfrac{adj}{hyp} * \dfrac{hyp}{opp} \\[5ex]
\therefore \cot \theta = \dfrac{adj}{opp}
$
Summary: The Unit Circle
| $\theta$ in DEG | $\theta$ in RAD | $\sin \theta$ | $\cos \theta$ | $\tan \theta$ | $\csc \theta$ | $\sec \theta$ | $\cot \theta$ |
|---|---|---|---|---|---|---|---|
| $0$ | $0$ | $0$ | $1$ | $0$ | $undefined$ | $1$ | $undefined$ |
| $30$ | $\dfrac{\pi}{6}$ | $\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{3}}{3}$ | $2$ | $\dfrac{2\sqrt{3}}{3}$ | $\sqrt{3}$ |
| $45$ | $\dfrac{\pi}{4}$ | $\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $1$ | $\sqrt{2}$ | $\sqrt{2}$ | $1$ |
| $60$ | $\dfrac{\pi}{3}$ | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $\sqrt{3}$ | $\dfrac{2\sqrt{3}}{3}$ | $2$ | $\dfrac{\sqrt{3}}{3}$ |
| $90$ | $\dfrac{\pi}{2}$ | $1$ | $0$ | $undefined$ | $1$ | $undefined$ | $0$ |
| $120$ | $\dfrac{2\pi}{3}$ | $\dfrac{\sqrt{3}}{2}$ | $-\dfrac{1}{2}$ | $-\sqrt{3}$ | $\dfrac{2\sqrt{3}}{3}$ | $-2$ | $-\dfrac{\sqrt{3}}{3}$ |
| $135$ | $\dfrac{3\pi}{4}$ | $\dfrac{\sqrt{2}}{2}$ | $-\dfrac{\sqrt{2}}{2}$ | $-1$ | $\sqrt{2}$ | $-\sqrt{2}$ | $-1$ |
| $150$ | $\dfrac{5\pi}{6}$ | $\dfrac{1}{2}$ | $-\dfrac{\sqrt{3}}{2}$ | $-\dfrac{\sqrt{3}}{3}$ | $2$ | $-\dfrac{2\sqrt{3}}{3}$ | $-\sqrt{3}$ |
| $180$ | $\pi$ | $0$ | $-1$ | $0$ | $undefined$ | $-1$ | $undefined$ |
| $210$ | $\dfrac{7\pi}{6}$ | $-\dfrac{1}{2}$ | $-\dfrac{\sqrt{3}}{2}$ | $\dfrac{\sqrt{3}}{3}$ | $-2$ | $-\dfrac{2\sqrt{3}}{3}$ | $\sqrt{3}$ |
| $225$ | $\dfrac{5\pi}{4}$ | $-\dfrac{\sqrt{2}}{2}$ | $-\dfrac{\sqrt{2}}{2}$ | $1$ | $-\sqrt{2}$ | $-\sqrt{2}$ | $1$ |
| $240$ | $\dfrac{4\pi}{3}$ | $-\dfrac{\sqrt{3}}{2}$ | $-\dfrac{1}{2}$ | $\sqrt{3}$ | $-\dfrac{2\sqrt{3}}{3}$ | $-2$ | $\dfrac{\sqrt{3}}{3}$ |
| $270$ | $\dfrac{3\pi}{2}$ | $-1$ | $0$ | $undefined$ | $-1$ | $undefined$ | $0$ |
| $315$ | $\dfrac{7\pi}{4}$ | $-\dfrac{\sqrt{2}}{2}$ | $\dfrac{\sqrt{2}}{2}$ | $-1$ | $-\sqrt{2}$ | $\sqrt{2}$ | $-1$ |
| $300$ | $\dfrac{5\pi}{3}$ | $-\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{2}$ | $-\sqrt{3}$ | $-\dfrac{2\sqrt{3}}{3}$ | $2$ | $-\dfrac{\sqrt{3}}{3}$ |
| $330$ | $\dfrac{11\pi}{6}$ | $-\dfrac{1}{2}$ | $\dfrac{\sqrt{3}}{2}$ | $-\dfrac{\sqrt{3}}{3}$ | $-2$ | $\dfrac{2\sqrt{3}}{3}$ | $-\sqrt{3}$ |
| $360$ | $2\pi$ | $0$ | $1$ | $0$ | $undefined$ | $1$ | $undefined$ |
Trigonometric Identities
\begin{array}{c | c} II & I \\ \hline III & IV \end{array} = \begin{array}{c | c} S & A \\ \hline T & C \end{array} = \begin{array}{c | c} Sine\:\: is\:\: positive & All\:\: are\:\: positive \\ \hline Tangent\:\: is\:\: positive & Cosine\:\: is\:\: positive \end{array}
The pneumonic to remember it:
First Quadrant to Fourth Quadrant (clockwise direction): A-C-T-S (ACTS of the Apostles)
First Quadrant to Fourth Quadrant (counter-clockwise direction): A-S-T-C: ADD - SUGAR - TO - COFFEE
Fourth Quadrant to Third Quadrant (counter-clockwise direction): C-A-S-T (Simon Peter, CAST your net into the
sea.)
Second Quadrant to Third Quadrant (clockwise direction): S-A-C-T
Cofunction Identities (Identities of Complements)
First Quadrant Identities
First Quadrant: All (sine, cosine, tangent) are positive
This implies that cosecant, secant, and cotangent are also positive.
Given: two angles say: $\alpha$ and $\beta$;
If $\alpha$ and $\beta$ are complementary, then the:
Sine function and the Cosine functions are cofunctions
$
\sin \alpha = \cos \beta \\[3ex]
\cos \alpha = \sin \beta \\[3ex]
$
Tangent function and the Cotangent functions are cofunctions
$
\tan \alpha = \cot \beta \\[3ex]
\cot \alpha = \tan \beta \\[3ex]
$
Secant function and the Cosecant functions are cofunctions
$
\sec \alpha = \csc \beta \\[3ex]
\csc \alpha = \sec \beta \\[3ex]
$
Given: one angle say: $\theta$;
First Quadrant Identities or Cofunction Identities or Identities of Complements
$0 \lt \theta \lt 90 ...Angle\:\: in\:\: Degrees \\[3ex]$
Reference Angle = $\theta$ ... Angle in Degrees
$0 \lt \theta \lt \dfrac{\pi}{2} ...Angle\:\: in\:\: Radians \\[5ex]$
Reference Angle = $\theta$ ... Angle in Radians
First Quadrant: sine, cosine, tangent are positive
This implies that cosecant, secant, and cotangent are also positive
Complement of $\theta$ = $90 - \theta$ where $\theta$ is in degrees:
Complement of $\theta$ = $\dfrac{\pi}{2} - \theta$ where $\theta$ is in radians:
$
(1.)\:\: \sin \theta = \cos (90 - \theta) \\[3ex]
(2.)\:\: \sin \theta = \cos \left(\dfrac{\pi}{2} - \theta \right) \\[5ex]
(3.)\:\: \cos \theta = \sin (90 - \theta) \\[3ex]
(4.)\:\: \cos \theta = \sin \left(\dfrac{\pi}{2} - \theta \right) \\[5ex]
(5.)\:\: \tan \theta = \cot (90 - \theta) \\[3ex]
(6.)\:\: \tan \theta = \cot \left(\dfrac{\pi}{2} - \theta \right) \\[5ex]
(7.)\:\: \cot \theta = \tan (90 - \theta) \\[3ex]
(8.)\:\: \cot \theta = \tan \left(\dfrac{\pi}{2} - \theta \right) \\[5ex]
(9.)\:\: \sec \theta = \csc (90 - \theta) \\[3ex]
(10.)\:\: \sec \theta = \csc \left(\dfrac{\pi}{2} - \theta \right) \\[5ex]
(11.)\:\: \csc \theta = \sec (90 - \theta) \\[3ex]
(12.)\:\: \csc \theta = \sec \left(\dfrac{\pi}{2} - \theta \right) \\[7ex]
$
Second Quadrant Identities or Identities of Supplements
$90 \lt \theta \lt 180$ ... Angle in Degrees
Reference Angle = $180 - \theta$ ... Angle in Degrees
$\dfrac{\pi}{2} \lt \theta \lt \pi$ ... Angle in Radians
Reference Angle = $\pi - \theta$ ... Angle in Radians
Second Quadrant: sine is positive
This implies that cosecant is also positive
Symmetric across the y-axis
Given: one angle say: $\theta$;
Supplement of $\theta$ = $180 - \theta$ where $\theta$ is in degrees:
Supplement of $\theta$ = $\pi - \theta$ where $\theta$ is in radians:
$
(1.)\;\; \sin \theta = \sin (180 - \theta) \\[3ex]
(2.)\;\; \sin (180 - \theta) = \sin \theta \\[3ex]
(3.)\;\; \sin \theta = \sin (\pi - \theta) \\[3ex]
(4.)\;\; \sin (\pi - \theta) = \sin \theta \\[3ex]
(5.)\;\; \cos \theta = -\cos (180 - \theta) \\[3ex]
(6.)\;\; \cos (180 - \theta) = -\cos \theta \\[3ex]
(7.)\;\; \cos \theta = -\cos (\pi - \theta) \\[3ex]
(8.)\;\; \cos (\pi - \theta) = -\cos \theta \\[3ex]
(9.)\;\; \tan \theta = -\tan (180 - \theta) \\[3ex]
(10.)\;\; \tan (180 - \theta) = -\tan \theta \\[3ex]
(11.)\;\; \tan \theta = -\tan (\pi - \theta) \\[3ex]
(12.)\;\; \tan (\pi - \theta) = -\tan \theta \\[5ex]
$
Third Quadrant Identities
$180 \lt \theta \lt 270$ ... Angle in Degrees
Reference Angle = $\theta - 180$ ... Angle in Degrees
$\pi \lt \theta \lt \dfrac{3\pi}{2}$ ... Angle in Radians
Reference Angle = $\theta - \pi$ ... Angle in Radians
Third Quadrant: tangent is positive
This implies that cotangent is also positive
Symmetric across the origin
Given: one angle say: $\theta$;
$
(1.)\;\; \sin \theta = -\sin (\theta - 180) \\[3ex]
(2.)\;\; \sin (\theta - 180) = -\sin \theta \\[3ex]
(3.)\;\; \sin \theta = -\sin (\theta - \pi) \\[3ex]
(4.)\;\; \sin (\theta - \pi) = -\sin \theta \\[3ex]
(5.)\;\; \cos \theta = -\cos (\theta - 180) \\[3ex]
(6.)\;\; \cos (\theta - 180) = -\cos \theta \\[3ex]
(7.)\;\; \cos \theta = -\cos (\theta - \pi) \\[3ex]
(8.)\;\; \cos (\theta - \pi) = -\cos \theta \\[3ex]
(9.)\;\; \tan \theta = \tan (\theta - 180) \\[3ex]
(10.)\;\; \tan (\theta - 180) = \tan \theta \\[3ex]
(11.)\;\; \tan \theta = \tan (\theta - \pi) \\[3ex]
(12.)\;\; \tan (\theta - \pi) = \tan \theta \\[5ex]
$
But if $\theta \lt 180$, use $\theta + 180$
Fourth Quadrant Identities
$270 \lt \theta \lt 360$ ... Angle in Degrees
Reference Angle = $360 - \theta$ ... Angle in Degrees
$\dfrac{3\pi}{2} \lt \theta \lt 2\pi$ ... Angle in Radians
Reference Angle = $2\pi - \theta$ ... Angle in Radians
Fourth Quadrant: cosine is positive
This implies that secant is also positive
Symmetric across the x-axis
Given: one angle say: $\theta$;
$
(1.)\;\; \sin \theta = -\sin (360 - \theta) \\[3ex]
(2.)\;\; \sin (360 - \theta) = -\sin \theta \\[3ex]
(3.)\;\; \sin \theta = -\sin (2\pi - \theta) \\[3ex]
(4.)\;\; \sin (2\pi - \theta) = -\sin \theta \\[3ex]
(5.)\;\; \cos \theta = \cos (360 - \theta) \\[3ex]
(6.)\;\; \cos (360 - \theta) = \cos \theta \\[3ex]
(7.)\;\; \cos \theta = \cos (2\pi - \theta) \\[3ex]
(8.)\;\; \cos (2\pi - \theta) = \cos \theta \\[3ex]
(9.)\;\; \tan \theta = -\tan (360 - \theta) \\[3ex]
(10.)\;\; \tan (360 - \theta) = -\tan \theta \\[3ex]
(11.)\;\; \tan \theta = -\tan (2\pi - \theta) \\[3ex]
(12.)\;\; \tan (2\pi - \theta) = -\tan \theta \\[3ex]
$
Reciprocal Identities
$
(1.)\;\; \csc \theta = \dfrac{1}{\sin \theta} \\[3ex]
(2.)\;\; \sec \theta = \dfrac{1}{\cos \theta} \\[3ex]
(3.)\;\; \cot \theta = \dfrac{1}{\tan \theta} \\[3ex]
$
From Reciprocal Identities
$
(1.)\;\; \sin \theta = \dfrac{1}{\csc \theta} \\[3ex]
(2.)\;\; \cos \theta = \dfrac{1}{\sec \theta} \\[3ex]
(3.)\;\; \tan \theta = \dfrac{1}{\cot \theta} \\[5ex]
$
Quotient Identities
$
(1.)\;\; \tan \theta = \dfrac{\sin \theta}{\cos \theta} \\[3ex]
(2.)\;\; \cot \theta = \dfrac{\cos \theta}{\sin \theta} \\[3ex]
$
As you can see, $\cot \theta$ has two formulas
$
\cot \theta = \dfrac{1}{\tan \theta} \\[5ex]
\cot \theta = \dfrac{\cos \theta}{\sin \theta} \\[5ex]
$
Sometimes, we shall use the first one. Sometimes, we shall use the second one.
We have to use the formula that will not give us an undefined value (where the denominator is zero)
Even Identities
Recall: a function, say $f(x)$ is even if $f(-x) = f(x)$
In Trigonometry, the cosine function and the secant function are even functions.
$
(1.)\;\; \cos (-\theta) = \cos \theta \\[3ex]
(2.)\;\; \sec (-\theta) = \sec \theta \\[5ex]
$
Odd Identities
Recall: a function, say $f(x)$ is odd if $f(-x) = -f(x)$
In Trigonometry, the sine function, the cosecant function,
the tangent function, and the cotangent function are odd functions.
$
(1.)\;\; \sin (-\theta) = -\sin \theta \\[3ex]
(2.)\;\; \csc (-\theta) = -\csc \theta \\[3ex]
(3.)\;\; \tan (-\theta) = -\tan \theta \\[3ex]
(4.)\;\; \cot (-\theta) = -\cot \theta \\[5ex]
$
Pythagorean Identities
$
(1.)\;\; \sin^2 \theta + \cos^2 \theta = 1 \\[3ex]
(2.)\;\; \tan^2 \theta + 1 = \sec^2 \theta \\[3ex]
(3.)\;\; \cot^2 \theta + 1 = \csc^2 \theta \\[3ex]
$
From Pythagorean Identities
$
(1.)\;\; \sin \theta = \pm \sqrt{1 - \cos^2 \theta} \\[3ex]
(2.)\;\; \cos \theta = \pm \sqrt{1 - \sin^2 \theta} \\[3ex]
(3.)\;\; \tan \theta = \pm \sqrt{\sec^2 \theta - 1} \\[3ex]
(4.)\;\; \sec \theta = \pm \sqrt{\tan^2 \theta + 1} \\[3ex]
(5.)\;\; \cot \theta = \pm \sqrt{\csc^2 \theta - 1} \\[3ex]
(6.)\;\; \csc \theta = \pm \sqrt{\cot^2 \theta + 1}
$
Trigonometric Formulas
Sum and Difference Formulas
$
(1.)\;\; \sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\[3ex]
(2.)\;\; \cos (\alpha \pm \beta) = \cos \alpha \cos \beta \mp \sin \alpha \sin \beta \\[3ex]
(3.)\;\; \tan (\alpha \pm \beta) = \dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta} \\[5ex]
$
Half-Angle Formulas
$
(1.)\;\; \sin \dfrac{\theta}{2} = \pm \sqrt{\dfrac{1 - \cos \theta}{2}} \\[5ex]
(2.)\;\; \cos {\theta \over 2} = \pm \sqrt{\dfrac{1 + \cos \theta}{2}} \\[5ex]
(3.)\;\; \tan {\theta \over 2} = \pm \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}} \\[5ex]
(4.)\;\; \tan {\theta \over 2} = \dfrac{\sin \theta}{1 + \cos \theta} \\[5ex]
(5.)\;\; \tan {\theta \over 2} = \dfrac{1 - \cos \theta}{\sin \theta} \\[5ex]
$
Formulas from Half-Angle Formulas
$
(1.)\;\; \sin^2 \dfrac{\theta}{2} = \dfrac{1 - \cos \theta}{2} \\[5ex]
(2.)\;\; \cos^2 \dfrac{\theta}{2} = \dfrac{1 + \cos \theta}{2} \\[5ex]
(3.)\;\; \tan^2 \dfrac{\theta}{2} = \dfrac{1 - \cos \theta}{1 + \cos \theta} \\[7ex]
$
Double-Angle Formulas
$
(1.)\;\; \sin (2\theta) = 2 \sin \theta \cos \theta \\[3ex]
(2.)\;\; \cos (2\theta) = \cos^2 \theta - \sin^2 \theta \\[3ex]
(3.)\;\; \cos (2\theta) = 1 - 2\sin^2 \theta \\[3ex]
(4.)\;\; \cos (2\theta) = 2\cos^2 \theta - 1 \\[3ex]
(5.)\;\; \tan (2\theta) = \dfrac{2\tan \theta}{1 - \tan^2 \theta} \\[5ex]
$
Formulas from Double-Angle Formulas
$
(1.)\;\; \sin^2 \theta = \dfrac{1 - \cos(2\theta)}{2} \\[5ex]
(2.)\;\; \cos^2 \theta = \dfrac{1 + \cos(2\theta)}{2} \\[5ex]
(3.)\;\; \tan^2 \theta = \dfrac{1 - \cos(2\theta)}{1 + \cos(2\theta)} \\[7ex]
$
Triple-Angle Formulas
$
(1.)\;\; \sin (3\theta) = 3 \sin \theta - 4\sin^3 \theta \\[3ex]
(2.)\;\; \cos (3\theta) = 4 \cos^3 \theta - 3 \cos \theta \\[3ex]
(3.)\;\; \tan (3\theta) = \dfrac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta} \\[7ex]
$
Sum-to-Product Formulas
$
(1.)\;\; \sin \alpha + \sin \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha -
\beta}{2}\right) \\[5ex]
(2.)\;\; \sin \alpha - \sin \beta = 2 \sin \left(\dfrac{\alpha - \beta}{2}\right) \cos \left(\dfrac{\alpha +
\beta}{2}\right) \\[5ex]
(3.)\;\; \cos \alpha + \cos \beta = 2 \cos \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha -
\beta}{2}\right) \\[5ex]
(4.)\;\; \cos \alpha - \cos \beta = -2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\alpha -
\beta}{2}\right) \\[5ex]
$
Ask students to write the compact form / shortened form of the first two Sum-to-Product Formulas.
Sum-to-Product Formulas (Compact Form of the First Two Formulas)
$
(1.)\;\; \sin \alpha \pm \sin \beta = 2 \sin \dfrac{\alpha \pm \beta}{2} \cos \dfrac{\alpha \mp \beta}{2}
\\[7ex]
$
Product-to-Sum Formulas
$
(1.) \sin \alpha * \sin \beta = \dfrac{1}{2} [\cos(\alpha - \beta) - \cos(\alpha + \beta)] \\[5ex]
(2.) \cos \alpha * \cos \beta = \dfrac{1}{2} [\cos(\alpha - \beta) + \cos(\alpha + \beta)] \\[5ex]
(3.) \sin \alpha * \cos \beta = \dfrac{1}{2} [\sin(\alpha + \beta) + \sin(\alpha + \beta)] \\[5ex]
$
Factoring Formulas
x is any trigonometric ratio
y is any trigonometric ratio
$
\underline{Difference\;\;of\;\;Two\;\;Squares} \\[3ex]
(1.)\;\;x^2 - y^2 = (x + y)(x - y) \\[5ex]
\underline{Difference\;\;of\;\;Two\;\;Cubes} \\[3ex]
(2.)\;\; x^3 - y^3 = (x - y)(x^2 + xy + y^2) \\[5ex]
\underline{Sum\;\;of\;\;Two\;\;Cubes} \\[3ex]
(3.)\;\; x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\[5ex]
$
Triangle Laws
Sine Law:
Applies to all triangles.
It states that the ratio of the side length of any triangle to the sine of the angle measure opposite that side
is the same for the three sides of the triangle.
$
\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} \\[5ex]
$
OR
The ratio of the angle measure of any triangle to the side length opposite that angle is the same for the three
angles of the triangle.
$
\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c} \\[5ex]
$
Cosine Law:
Applies to all triangles.
It states that the square of a side of a triangle is the difference between the sum of the squares of the other
two sides and twice the product of the two sides and the included angle.
$
a^2 = b^2 + c^2 - 2bc \cos A \\[3ex]
\cos A = \dfrac{b^2 + c^2 - a^2}{2bc} \\[5ex]
\rightarrow A = \cos^{-1} \left(\dfrac{b^2 + c^2 - a^2}{2bc}\right) \\[5ex]
b^2 = a^2 + c^2 - 2ac \cos B \\[3ex]
\cos B = \dfrac{a^2 + c^2 - b^2}{2ac} \\[5ex]
\rightarrow B = \cos^{-1} \left(\dfrac{a^2 + c^2 - b^2}{2ac}\right) \\[5ex]
c^2 = a^2 + b^2 - 2ab \cos C \\[3ex]
\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} \\[5ex]
\rightarrow C = \cos^{-1} \left(\dfrac{a^2 + b^2 - c^2}{2ab}\right) \\[5ex]
$
Theorems
(1.) Sum of Angles of a Triangle Theorem:
The sum of the interior angles of a triangle is 180°
(2.) In a 30° – 60° – 90° right triangle; the length of the hypotenuse
is twice the length of the short side.
(3.) In a 30° – 60° – 90° right triangle; the length of the middle side
is the square root of 3 times the length of short side.
(4.) In a 45° – 45° – 90° right triangle theorem (Right Isosceles Triangle
Theorem); the length of the hypotenuse is the square root of 2 times the length of either side.
(5.) Pythagorean Theorem: Applies only to right triangles.
In a right triangle, the square of the length of the hypotenuse is the
sum of the squares of the short side and the middle side.
$hyp^2 = leg^2 + leg^2$
(6.) Converse of the Pythagorean Theorem (Right Triangle Theorem):
If the square of the long side(hypotenuse) is the
sum of the squares of the other two sides, then the triangle is a right triangle.
(7.) Acute Triangle Theorem: If the square of the long side is less than the sum of the squares of the
other two sides, then the triangle is an acute triangle.
(8.) Obtuse Triangle Theorem: If the square of the long side is greater than the sum of the squares of
the other two sides, then the triangle is an obtuse triangle.
(9.) Triangle Inequality Theorem: This is the theorem that determines if you can form a triangle using
any three lengths.
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
(10.) Side Length – Angle Measure Theorem: If any two side lengths of a triangle are unequal; the
angles of the triangle are also unequal, and the measure of an angle is opposite the length of the side facing
that angle as regards size.
The measure of the smallest angle is opposite the shortest side length.
The measure of the greatest angle is opposite the longest side length.
The measure of the middle angle is opposite the middle side length.
In other words; regarding size, the measure of an angle is opposite the length of the side facing the
angle; or the side length facing an angle is opposite the angle measure as regards size.
Small side faces small angle, middle side faces middle angle, big side faces big angle
(11.) Exterior Angle of a Triangle Theorem: The exterior angle of a triangle is the sum of the two
interior opposite angles.
(12.) Isosceles Base Angles Theorem: The base angles of an isosceles triangle are equal.
(13.) Converse of the Isosceles Base Angles Theorem: If the base angles of a triangle are equal, then the
triangle is an isosceles triangle.
(14.) Perpendicular Bisector of the Base of an Isosceles Triangle Theorem: It states that if a line
bisects the vertex angle of an isosceles triangle, then the line is also the perpendicular
bisector of the base (the line also bisects the base of that isosceles triangle at right angles).
(15.) Perpendicular Height to Base of Isosceles Triangle Theorem It states that the perpendicular height
drawn from the apex of an isosceles triangle to the base:
(a.) bisects the base
(b.) bisects the apex angle.
(16.) Side-Splitter Theorem Applies to all triangles with inserted parallel lines as applicable.
It states that if a line segment is parallel to one side of a triangle and intersects the other two sides of
the triangle, then it divides those two sides proportionally.
(17.) Midpoint Theorem Applies to all triangles in which a line segment
joins the midpoints of any two sides of the triangle.
It states that if a line segment joins any two sides of a triangle, then the line segment:
(a.) is parallel to the third side.
(b.) bisects the third side.
(18.) Converse of the Midpoint Theorem A line segment drawn through the midpoint of one side of a
triangle and parallel to another side, bisects the third side.
(19.) Scale Factor, Perimeter Ratio, and Area Ratio of Similar Figures Theorem Applies to all similar
figures including similar triangles.
It states that for two similar figures, the ratio of the perimeters is the same as the scale factor; and the
ratio of the areas is the ratio of the square of the scale factor.
Circle Formulas
Except stated otherwise, use:
$
d = diameter \\[3ex]
r = radius \\[3ex]
L = arc\:\:length \\[3ex]
A = area\;\;of\;\;sector \\[3ex]
P = perimeter\;\;of\;\;sector \\[3ex]
\theta = central\:\:angle \\[3ex]
\pi = \dfrac{22}{7} \\[5ex]
RAD = radians \\[3ex]
^\circ = DEG = degrees \\[7ex]
\underline{\theta\;\;in\;\;DEG} \\[3ex]
L = \dfrac{2\pi r\theta}{360} \\[5ex]
\theta = \dfrac{180L}{\pi r} \\[5ex]
r = \dfrac{180L}{\pi \theta} \\[5ex]
A = \dfrac{\pi r^2\theta}{360} \\[5ex]
\theta = \dfrac{360A}{\pi r^2} \\[5ex]
r = \dfrac{360A}{\pi\theta} \\[5ex]
P = \dfrac{r(\pi\theta + 360)}{180} \\[7ex]
\underline{\theta\;\;in\;\;RAD} \\[3ex]
L = r\theta \\[5ex]
\theta = \dfrac{L}{r} \\[5ex]
r = \dfrac{L}{\theta} \\[5ex]
A = \dfrac{r^2\theta}{2} \\[5ex]
\theta = \dfrac{2A}{r^2} \\[5ex]
r = \sqrt{\dfrac{2A}{\theta}} \\[5ex]
P = r(\theta + 2) \\[7ex]
Circumference\:\:of\:\:a\:\:circle = 2\pi r = \pi d \\[3ex]
L = \dfrac{2A}{r} \\[5ex]
r = \dfrac{2A}{L} \\[5ex]
A = \dfrac{Lr}{2}
$
In the diagram, EFH is a triangle, $|\overline{EG}|$ = 50cm and $|\overline{GH}|$ = x cm.
Find, correct to one decimal place,:
(a.) the value of x
(b.) $|\overline{FH}|$
$ (a.) \\[3ex] \underline{\triangle GFH} \\[3ex] \angle GFH + \angle GHF + \angle FGH = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[3ex] \angle GFH + 90 + 70 = 180 \\[3ex] \angle GFH = 180 - 90 - 70 \\[3ex] \angle GFH = 20^\circ \\[3ex] \underline{\triangle EFG} \\[3ex] \angle EGF = \angle GHF + \angle GFH ...exterior\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle EGF = 90 + 20 \\[3ex] \angle EGF = 110^\circ \\[3ex] \angle EFG + \angle FEG + \angle EGF = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[3ex] \angle EFG + 30 + 110 = 180 \\[3ex] \angle EFG = 180 - 30 - 110 \\[3ex] \angle EFG = 40^\circ \\[3ex] \dfrac{|FG|}{\sin \angle FEG} = \dfrac{|EG|}{\sin \angle EFG} ...Sine\;\;Law \\[5ex] \dfrac{|FG|}{\sin 30} = \dfrac{|EG|}{\sin 40} ...Sine\;\;Law \\[5ex] |FG| = \dfrac{50\sin 30}{\sin 40} \\[5ex] |FG| = 38.89309567\;cm \\[3ex] \underline{\triangle GFH} \\[3ex] \dfrac{|GH|}{\sin \angle GFH} = \dfrac{|FG|}{\sin \angle GHF} ...Sine\;\;Law \\[5ex] \dfrac{x}{\sin 20} = \dfrac{38.89309567}{\sin 90} \\[5ex] x = \dfrac{38.89309567 \sin 20}{\sin 90} \\[5ex] x = 13.30222216 \\[3ex] x \approx 13.3\;cm...to\;\;1\;\;d.p \\[5ex] (b.) \\[3ex] \underline{\triangle GFH} \\[3ex] |FG|^2 = |FH|^2 + |GH|^2 ...Pythagorean\;\;Theorem \\[4ex] (38.89309567)^2 = |FH|^2 + (13.30222216)^2 \\[3ex] |FH|^2 = 1512.672891 - 176.9491144 \\[3ex] |FH|^2 = 1335.723777 \\[3ex] |FH| = \sqrt{1335.723777} \\[3ex] |FH| = 36.547555 \\[3ex] |FH| \approx 36.5\;cm...to\;\;1\;\;d.p $
If Dogbe walked 5 km and Mamu 3 km from the cottage tk their farms, find, correct to:
(i.) two significant figures, the distance between the two farms.
(ii.) the nearest degree, the bearing of Mamu's farm from Dogbe's.
(b.) A ladder 10 m long leaned against a vertical wall x m high.
The distance between the wall and the foot of the ladder is 2 m longer than the height of the wall.
Calculate the value of x.
(a.)
Let:
Dogbe's farm = D
Mamu's farm = M
Cottage = C
Let us represent the information diagrammatically
$ \alpha = \angle MDC \\[3ex] \underline{\text{Right Triangle MDC}} \\[3ex] (i.) \\[3ex] c^2 = 5^2 + 3^2 ...\text{Pythagorean Theorem} \\[3ex] c = \sqrt{25 + 9} \\[3ex] c = 5.830951895 \\[3ex] c \approx 5.8\;km...\text{to two significant figures} \\[5ex] \tan \alpha = \dfrac{3}{5}...\text{SOHCAHTOA} \\[5ex] \alpha = \tan^{-1}(0.6) \\[3ex] \alpha = 30.96375653 \\[3ex] (ii.) \\[3ex] \text{Bearing of M from D} \\[3ex] = 200 + \alpha \\[3ex] = 200 + 30.96375653 \\[3ex] = 230.96375653 \\[3ex] \approx 231^\circ ...\text{to the nearest degree} \\[3ex] $ (b.)
Let us represent the information diagrammatically.
$ hyp^2 = leg^2 + leg^2 ...\text{Pythagorean Theorem} \\[3ex] 10^2 = x^2 + (2 + x)^2 \\[3ex] 100 = x^2 + (2 + x)(2 + x) \\[3ex] 100 = x^2 + 4 + 2x + 2x + x^2 \\[3ex] 100 = 2x^2 + 4x + 4 \\[3ex] 2x^2 + 4x + 4 - 100 = 0 \\[3ex] 2x^2 + 4x - 96 = 0 \\[3ex] x^2 + 2x - 48 = 0 \\[3ex] (x + 8)(x - 6) = 0 \\[3ex] x + 8 = 0 \hspace{2em}OR\hspace{2em} x - 6 = 0 \\[3ex] x = -8...\text{discard because the length cannot be negative} \\[3ex] OR \\[3ex] x = 6\;m $
Musa is standing 8 km due west of the tree.
(i.) Illustrate the information on a diagram.
(ii.) How far is Musa from the building?
(iii.) Find the bearing of Musa from the building.
Let:
Building = B
Tree = T
Musa = M
(i.) The diagram is:
$ (ii.) \\[3ex] \underline{\triangle MBT} \\[3ex] |MB|^2 = |BT|^2 + |MT|^2...\text{Pythagorean Theorem} \\[3ex] = 8^2 + 8^2 \\[3ex] = 64 + 64 \\[3ex] = 128 \\[3ex] |MB| = \sqrt{128} \\[3ex] |MB| = 11.3137085\;km \\[3ex] $ Let us update the diagram.
$ \tan \alpha = \dfrac{opp}{adj} ...SOHCAHTOA \\[5ex] \tan \alpha = \dfrac{8}{8} \\[5ex] \tan \alpha = 1 \\[3ex] \alpha = \tan^{-1}(1) \\[3ex] \alpha = 45^\circ \\[5ex] (ii.) \\[3ex] \text{Bearing of Musa from the Building} = 180^\circ + \alpha \\[3ex] = 180 + 45 \\[3ex] = 225^\circ $
In the diagram, PQR is an equilateral triangle of side 18 cm.
M is the midpoint of $\overline{QR}$
An arc of a circle with centre P touches $\overline{QR}$ at M and meets $\overline{PQ}$ at A and $\overline{PR}$ at B
Calculate, correct to two decimal places, the area of the shaded region.
$\left[\text{Take } \pi = \dfrac{22}{7}\right]$
Construction: Draw a perpendicular line from P to M
|PM| is the radius, r of the circle...centre P to circumference, M
|PM| is also the perpendicular height, h of the triangle
θ = Angle subtended at center of circle = 60° ...angle in the equilateral triangle PQR
Area of the shaded region = Area of Triangle PQR − Area of Sector PAB
$ \underline{\triangle PQR} \\[3ex] 18^2 = h^2 + 9^2 ...\text{Pythagorean Theorem} \\[4ex] 324 = h^2 + 81 \\[3ex] h^2 + 81 = 324 \\[3ex] h^2 = 324 - 81 \\[3ex] h^2 = 243 \\[3ex] h = \sqrt{243} \\[3ex] h = 15.58845727\;cm \\[5ex] \text{Area of Triangle PQR} = \dfrac{bh}{2} \\[5ex] = \dfrac{18(15.58845727)}{2} \\[5ex] = 140.2961154\;cm^2 \\[5ex] \underline{\text{Area of Sector PAB}} \\[3ex] \text{Area of Sector} = \dfrac{\pi r^2 \theta}{360} \\[5ex] = \dfrac{22}{7} * 15.58845727^2 * \dfrac{60}{360} \\[5ex] = 127.2857143\;cm^2 \\[5ex] \text{Area of the shaded region} \\[3ex] = 140.2961154 - 127.2857143 \\[3ex] = 13.01040108 \\[3ex] \approx 13.01\;cm^2 ...\text{to 2 decimal places} $
Another town, T is 8km from K on a bearing 155°
Calculate:
(i.) to the nearest kilometer, the distance of T from J
(ii.) to the nearest degree, the bearing of T from J
Let us draw the diagram
θ, φ = angles
$ (i.) \\[3ex] Distance\;\;of\;\;T\;\;from\;\;J = k \\[3ex] \angle TKJ = 25 + 65 = 90^\circ \\[3ex] k^2 = j^2 + t^2 - 2jt \cos \angle TKJ...Cosine\;\;Law \\[3ex] k^2 = 8^2 + 20^2 - 2(8)(20) * \cos 90 \\[3ex] k^2 = 64 + 400 - 320 * 0 \\[3ex] k^2 = 464 - 0 \\[3ex] k^2 = 464 \\[3ex] k = \sqrt{464} \\[3ex] k = 21.54065923 \\[3ex] k \approx 22\;km \\[3ex] (ii.) \\[3ex] \dfrac{\sin \theta}{j} = \dfrac{\sin 90}{k}...Sine\;\;Law \\[5ex] \dfrac{\sin \theta}{8} = \dfrac{\sin 90}{21.54065923} \\[5ex] \sin \theta = \dfrac{8 * \sin 90}{21.54065923} \\[5ex] \sin \theta = \dfrac{8 * 1}{21.54065923} \\[5ex] \sin \theta = 0.3713906763 \\[3ex] \theta = \sin^{-1}{0.3713906763} \\[3ex] \theta = 21.80140948 \\[3ex] \theta + \phi = 65 ...alternate \angle s\;\;are\;\;equal \\[3ex] \phi = 65 - \theta \\[3ex] \phi = 65 - 21.80140948 \\[3ex] \phi = 43.19859052 \\[5ex] Bearing\;\;of\;\;T\;\;from\;\;J \\[3ex] = 180 + \phi \\[3ex] = 180 + 43.19859052 \\[3ex] = 223.1985905 \\[3ex] \approx 223^\circ $
$ \underline{\text{Unit Circle Trigonometry}} \\[3ex] \tan x = \sqrt{3} \\[3ex] x = \tan^{-1}(\sqrt{3}) \\[3ex] x = 60^\circ \\[3ex] \sin x = \sin 60^\circ = \dfrac{\sqrt{3}}{2} \\[5ex] \cos x = \cos 60^\circ = \dfrac{1}{2} \\[5ex] \implies \\[3ex] \dfrac{(\cos x)^2 - \sin x}{(\sin x)^2 + \cos x} \\[5ex] = \left[\left(\dfrac{1}{2}\right)^2 - \dfrac{\sqrt{3}}{2}\right] \div \left[\left(\dfrac{\sqrt{3}}{2}\right)^2 + \dfrac{1}{2}\right] \\[5ex] = \left(\dfrac{1}{4} - \dfrac{\sqrt{3}}{2}\right) \div \left(\dfrac{3}{4} + \dfrac{1}{2}\right) \\[5ex] = \dfrac{1 - 2\sqrt{3}}{4} \div \dfrac{3 + 2}{4} \\[5ex] = \dfrac{1 - 2\sqrt{3}}{4} * \dfrac{4}{5} \\[5ex] = \dfrac{1 - 2\sqrt{3}}{5} $
After spending some time with her friend, she continued to a nearby town, M, 3t km away on a bearing of 155°.
If the distance between K and M is $6\sqrt{13}$ km:
(a.) illustrate the information in a diagram;
(b.) calculate, correct to the nearest whole number, the:
(i.) value of t;
(ii.) bearing of M from K.
Let β be the angle opposite side 3t
(a.) The diagram is drawn as shown:
$ (b.)(i.) \\[3ex] \triangle KLM \text{ is a right triangle} \\[3ex] \underline{\text{Pythagorean Theorem}} \\[3ex] hyp^2 = leg^2 + leg^2 \\[3ex] (6\sqrt{3})^2 = (2t)^2 + (3t)^2 \\[3ex] 36(13) = 4t^2 + 9t^2 \\[3ex] 13t^2 = 36(13) \\[3ex] t^2 = \dfrac{36(13)}{13} \\[5ex] t^2 = 36 \\[3ex] t = \sqrt{36}...\text{side is a positive number} \\[3ex] \text{discard the negative value} \\[3ex] t = 6\;km \\[5ex] (ii.) \\[3ex] \underline{\text{SOHCAHTOA}} \\[3ex] \cos\beta = \dfrac{adj}{hyp} \\[5ex] = \dfrac{2t}{6\sqrt{13}} \\[5ex] = \dfrac{2(6)}{6\sqrt{13}} \\[5ex] = \dfrac{12}{6\sqrt{13}} \\[5ex] = \dfrac{2}{\sqrt{13}} \\[5ex] = 0.5547001962 \\[5ex] \beta = \cos^{-1}(0.5547001962) \\[3ex] = 56.30993247 \\[5ex] \text{Bearing of M from K} = 65 + \beta \\[3ex] = 65 + 56.30993247 \\[3ex] = 121.3099325 \\[3ex] \approx 121^\circ ...\text{to the nearest whole number} $
$ \cos(2y - 16) = \dfrac{1}{2} \\[5ex] 2y - 16 = \cos^{-1}\left(\dfrac{1}{2}\right) \\[5ex] 2y - 16 = 60 \\[3ex] 2y = 60 + 16 \\[3ex] 2y = 76 \\[3ex] y = \dfrac{76}{2} \\[5ex] y = 38^\circ \\[3ex] $ Check
| LHS | RHS |
|---|---|
| $ \cos(2y - 16) \\[3ex] \cos[2(38) - 16] \\[3ex] \cos(76 - 16) \\[3ex] \cos 60 \\[3ex] \dfrac{1}{2} $ | $\dfrac{1}{2}$ |
The residence of X and Y are 180 meters apart.
(a.) Illustrate the information in a diagram.
(b.) Find, correct to three significant figures, the:
(i.) bearing of X from Y
(ii.) distance between P and Y
$ From\;\;North\;\;of\;\;\;P\;\;to\;\;X: 57^\circ \\[3ex] Remaining:\; 90 - 57 = 33^\circ \\[3ex] From\;\;North\;\;of\;\;\;P\;\;to\;\;Y: 150^\circ \\[3ex] 150 - 90 = 60 \\[3ex] \implies \\[3ex] \angle P = 33 + 60 = 93^\circ \\[3ex] $ (a.) θ = angle
The diagram of the information is:
$ (b.)(i.) \\[3ex] \dfrac{\sin Y}{y} = \dfrac{\sin P}{p} ...Sine\;\;Law \\[5ex] \dfrac{\sin Y}{60} = \dfrac{\sin 93}{180} \\[5ex] Y = \sin^{-1}\left(\dfrac{60\sin 93}{180}\right) \\[5ex] Y = 19.44346134^\circ \\[3ex] \theta = 60^\circ ... alternate\;\;interior\;\angle s\;\;are\;\;congruent \\[3ex] Bearing\;\;of\;\;X\;\;from\;\;Y = 270 + \theta + \angle Y \\[3ex] = 270 + 60 + 19.44346134 \\[3ex] = 349.4434613 \\[3ex] \approx 349^\circ ...to\;\;3\;s.f \\[5ex] (ii.) \\[3ex] \angle X + \angle Y + \angle P = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[3ex] \angle X + 19.44346134 + 93 = 180 \\[3ex] \angle X + 112.4434613 = 180 \\[3ex] \angle X = 180 - 112.4434613 \\[3ex] \angle X = 67.55653866^\circ \\[3ex] |PY| = x ...diagram \\[3ex] x^2 = p^2 + y^2 - 2py \cos X...Cosine\;\;Law \\[3ex] x^2 = 180^2 + 60^2 - 2(180)(60) \cos 67.55653866 \\[3ex] x^2 = 32400 + 3600 - 21600 \cdot 0.381771575 \\[3ex] x^2 = 32400 + 3600 - 8246.26602 \\[3ex] x^2 = 27753.73398 \\[3ex] x = \sqrt{27753.73398} \\[3ex] x = 166.5945197 \\[3ex] x \approx 167\;m ...to\;\;3\;s.f $
$ \tan x = \dfrac{15}{8} = \dfrac{opp}{adj}...SOHCAHTOA \\[5ex] opp = 15 \\[3ex] adj = 8 \\[3ex] hyp^2 = opp^2 + adj^2 ...\text{Pythagorean Theorem} \\[3ex] hyp^2 = 15^2 + 8^2 \\[3ex] hyp^2 = 225 + 64 \\[3ex] hyp^2 = 289 \\[3ex] hyp = \sqrt{289} \\[3ex] hyp = 17 \\[5ex] \underline{SOHCAHTOA} \\[3ex] \sin x = \dfrac{opp}{hyp} = \dfrac{15}{17} \\[5ex] \cos x = \dfrac{adj}{hyp} = \dfrac{8}{17} \\[5ex] 5\sin x - 6\cos x \\[3ex] = 5\left(\dfrac{15}{17}\right) - 6\left(\dfrac{8}{17}\right) \\[5ex] = \dfrac{75}{17} - \dfrac{48}{17} \\[5ex] = \dfrac{75 - 48}{17} \\[5ex] = \dfrac{27}{17} $
$ \tan x = \dfrac{15}{8} \\[5ex] x = \tan^{-1}\left(\dfrac{15}{8}\right) \\[5ex] 5\sin x - 6\cos x \\[3ex] $
If |XY| = 19.5km and |YZ| = 11.5km
(a.) Illustrate the information in a diagram.
(b.) Calculate, correct to the nearest whole number,
(i.) ∠ZXY
(ii.) |XZ|
(a.) The diagram is:
$ (b.) \\[3ex] (i.) \\[3ex] \dfrac{\sin \angle ZXY}{|ZY|} = \dfrac{\sin \angle YZX}{|YX|}...Sine\;\;Law \\[5ex] \dfrac{\sin \theta}{11.5} = \dfrac{\sin 100}{19.5} \\[5ex] \sin \theta = \dfrac{11.5 * \sin 100}{19.5} \\[5ex] \sin \theta = \dfrac{11.5 * 0.984807753}{19.5} \\[5ex] \sin \theta = 0.5807840595 \\[3ex] \theta = \sin^{-1}(0.5807840595) \\[3ex] \theta = 35.50570812 \\[3ex] \theta \approx 36^\circ \\[3ex] \angle ZYX + \angle YZX + \angle ZXY = 180^\circ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle ZYX + 100 + 35.50570812 = 180 \\[3ex] \angle ZYX = 180 - 100 - 35.50570812 \\[3ex] \angle ZYX = 44.49429188^\circ \\[3ex] (ii.) \\[3ex] \dfrac{|XZ|}{\sin \angle ZYX} = \dfrac{|YX|}{\sin \angle YZX}...Sine\;\;Law \\[5ex] \dfrac{|XZ|}{\sin 44.49429188} = \dfrac{19.5}{\sin 100} \\[5ex] |XZ| = \dfrac{19.5 * \sin 44.49429188}{\sin 100} \\[5ex] |XZ| = \dfrac{19.5 * 0.7008382029}{0.984807753} \\[5ex] |XZ| = 13.66634496 \\[3ex] |XZ| \approx 14\;km $
The angles of elevation of the top, T, of the tree from X and Y on the horizontal ground with the foot of the tree are 43° and 38° respectively.
(i.) Illustrate the information in a diagram.
(ii.) Find, correct to one decimal place, the height of the tree.
(a.)
$ \text{height of the tree} = |TZ| = h \\[3ex] |YZ| = k \\[5ex] \underline{\text{Right Triangle TYZ}} \\[3ex] \tan 43^\circ = \dfrac{h}{k}...\text{SOHCAHTOA} \\[5ex] k = \dfrac{h}{\tan 43} ...eqn.(1) \\[5ex] \underline{\text{Right Triangle TXZ}} \\[3ex] \tan 38^\circ = \dfrac{h}{19 + k}...\text{SOHCAHTOA} \\[5ex] k + 19 = \dfrac{h}{\tan 38} \\[5ex] k = \dfrac{h}{\tan 38} - 19 ...eqn.(2) \\[5ex] k = k \implies eqn.(2) = eqn.(1) \\[3ex] \dfrac{h}{\tan 38} - 19 = \dfrac{h}{\tan 43} \\[5ex] \dfrac{h}{\tan 38} - \dfrac{h}{\tan 43} = 19 \\[5ex] \dfrac{h\tan 43 - h\tan 38}{(\tan 38)(\tan 43)} = 19 \\[5ex] h(\tan 43 - \tan 38) = 19(\tan 38)(\tan 43) \\[3ex] h = \dfrac{19(\tan 38)(\tan 43)}{\tan 43 - \tan 38} \\[5ex] h = 91.53409704 \\[3ex] h \approx 91.5\;m...\text{to one decimal place} $
In the diagram, MNR is right triangle.
$|\overline{MN}| = 15\;m$, $|\overline{MR}| = 10\sqrt{3}\;m$ and $\angle MQN = 72^\circ$.
Calculate, correct to the nearest whole number:
(a.) the value of the angle marked x;
(b.) $|\overline{QR}|$;
(c.) area of $\triangle MQR$
$ \theta = \angle YXZ \\[3ex] (a.) \\[3ex] \underline{\triangle MQN} \\[3ex] \angle QMN + 72^\circ + 90^\circ = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] \angle QMN = 180 - 72 - 90 \\[3ex] \angle QMN = 18^\circ \\[5ex] \underline{\triangle MRN} \\[3ex] \cos (18 + x) = \dfrac{adj}{hyp}...SOHCAHTOA \\[5ex] = \dfrac{15}{10\sqrt{3}} \\[5ex] = \dfrac{15}{10\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = \dfrac{\sqrt{3}}{2} \\[5ex] 18 + x = \cos^{-1}\left(\sqrt{3}{2}\right) \\[5ex] 18 + x = 30 \\[3ex] x = 30 - 18 \\[3ex] x = 12^\circ \\[5ex] (b.) \\[3ex] \angle MQR + 72^\circ = 180^\circ ...\text{sum of angles on a straight line} \\[3ex] \angle MQR = 180 - 72 \\[3ex] \angle MQR = 108^\circ \\[5ex] \underline{\triangle MQR} \\[3ex] \dfrac{|\overline{QR}|}{\sin \angle x} = \dfrac{10\sqrt{3}}{\sin \angle MQR}...\text{Sine Law} \\[5ex] \dfrac{|\overline{QR}|}{\sin 12^\circ} = \dfrac{10\sqrt{3}}{\sin \angle 108^\circ} \\[5ex] |\overline{QR}| = \dfrac{10\sqrt{3} * \sin 12^\circ}{\sin \angle 108^\circ} \\[5ex] |\overline{QR}| = 3.786458594 \\[3ex] |\overline{QR}| \approx 4\;m ...\text{to the nearest whole number} \\[5ex] (c.) \\[3ex] \underline{\triangle MQR} \\[3ex] \angle MRQ + \angle MQR + \angle RMQ = 180^\circ ...\text{sum of angles in a triangle} \\[3ex] \angle MRQ + 108^\circ + 12^\circ = 180^\circ \\[3ex] \angle MRQ = 180 - 108 - 12 \\[3ex] \angle MRQ = 60^\circ \\[5ex] \text{Area of Triangle MQR} \\[3ex] = \dfrac{1}{2} * |\overline{MR}| * |\overline{QR}| * \sin \angle MRQ \\[5ex] = \dfrac{1}{2} * 10\sqrt{3} * 3.786458594 * \sin 60^\circ \\[5ex] = 28.39843946 \\[3ex] \approx 28\;m^2 $
$ \dfrac{5}{3}\sin(x - 29^\circ) = \dfrac{1}{2} \\[5ex] \sin(x - 29) = \dfrac{1}{2} * \dfrac{3}{5} \\[5ex] \sin(x - 29) = \dfrac{3}{10} \\[5ex] x - 29 = \sin^{-1}(0.3) \\[3ex] x = \sin^{-1}(0.3) + 29 \\[3ex] x = 46.45760312 \\[3ex] x \approx 46^\circ ...\text{to the nearest degree} \\[3ex] $ Check
| LHS | RHS |
|---|---|
| $ \dfrac{5}{3}\sin(x - 29^\circ) \\[5ex] \dfrac{5}{3}\sin(46.45760312^\circ - 29^\circ) \\[5ex] \dfrac{5}{3} * \sin (17.45760312) \\[5ex] = \dfrac{5}{3} * 0.2999999999 \\[5ex] = \dfrac{1.5}{3} \\[5ex] = 0.5 \\[3ex] = \dfrac{1}{2} $ | $\dfrac{1}{2}$ |
Calculate, correct to:
(a.) two significant figures, $|YZ|$
(b.) the nearest degree, the bearing of Y from Z
Let us represent the information on a diagram
$ \phi = \angle XZY \\[3ex] \underline{\triangle ZXY} \\[3ex] (a.) \\[3ex] |YZ|^2 = 20^2 + 15^2 ...\text{Pythagorean Theorem} \\[3ex] |YZ| = \sqrt{400 + 225} \\[3ex] |YZ| = 25\;km \\[5ex] \tan \angle XZY = \dfrac{|XY|}{|XZ|}...\text{SOHCAHTOA} \\[5ex] \tan \phi = \dfrac{15}{20} \\[5ex] \phi = \tan^{-1}\left(\dfrac{15}{20}\right) \\[5ex] \phi = 36.86989765^\circ \\[3ex] (b.) \\[3ex] \text{Bearing of Y from Z} \\[3ex] = 90 + \phi \\[3ex] (b.) \\[3ex] = 90 + 36.86989765^\circ \\[3ex] = 126.86989765^\circ \\[3ex] \approx 127^\circ ...\text{to the nearest degree} $
It then sails 7km from Y on a bearing of 068° to a point Z
Calculate the:
(a.) distance XZ, correct to two decimal places;
(b.) bearing of Z from X, correct to the nearest degree.
Let us represent the information on a diagram
$ \alpha, \beta, \theta ...\text{ are angles in the triangle} \\[3ex] \beta = 37^\circ ... alternate\;\;interior\;\angle s\;\;are\;\;congruent \\[3ex] 68 + \alpha = 90^\circ ...Right\;\angle \\[3ex] \alpha = 90 - 68 \\[3ex] \alpha = 22^\circ \\[4ex] \angle Y = 37^\circ + 90^\circ + 22^\circ ...diagram \\[4ex] \angle Y = 149^\circ \\[5ex] (a.) \\[3ex] |XZ| = y ... diagram \\[3ex] y^2 = 6^2 + 7^2 - 2(6)(7) \cos 149^\circ ...Cosine\;\;Law \\[4ex] = 36 + 49 - 84 \cdot -0.8571673007 \\[3ex] = 85 + 72.00205326 \\[3ex] = 157.0020533 \\[3ex] y = \sqrt{157.0020533} \\[3ex] y = 12.53004602 \\[3ex] y \approx 12.53\;km...\text{to two decimal places} \\[5ex] \dfrac{\sin\theta}{7} = \dfrac{\sin 149^\circ}{y} ...Sine\;\;Law \\[6ex] \sin\theta = \dfrac{7\sin 149}{y} \\[5ex] \sin\theta = \dfrac{7(0.5150380749)}{12.53004602} \\[5ex] \sin\theta = 0.2877297113 \\[3ex] \theta = \sin^{-1}(0.2877297113) \\[4ex] \theta = 16.72208586^\circ \\[4ex] (b.) \\[3ex] \text{bearing of Z from X} = 37^\circ + \theta \\[4ex] = 37 + 16.72208586 \\[3ex] = 53.72208586 \\[3ex] \approx 54^\circ ...\text{to the nearest degree} $
$ \cos\theta = \dfrac{15}{17} = \dfrac{adj}{hyp} ...\text{SOHCAHTOA} \\[5ex] opp = ? \\[3ex] adj = 15 \\[3ex] hyp = 17 \\[3ex] opp^2 + adj^2 = hyp^2 ...\text{Pythagorean Theorem} \\[5ex] opp^2 + 15^2 = 17^2 \\[3ex] opp = \sqrt{17^2 - 15^2} \\[3ex] opp = 8 \\[3ex] \tan\theta = \dfrac{opp}{adj} = \dfrac{8}{15} ...\text{SOHCAHTOA} \\[5ex] 1 + 2\tan\theta \\[3ex] = 1 + 2\left(\dfrac{8}{15}\right) \\[5ex] = 1 + \dfrac{16}{15} \\[5ex] = \dfrac{15}{15} + \dfrac{16}{15} \\[5ex] = \dfrac{31}{15} \\[5ex] \implies \\[3ex] \dfrac{\tan\theta}{1 + 2\tan\theta} \\[5ex] = \tan\theta \div (1 + 2\tan\theta) \\[3ex] = \dfrac{8}{15} \div \dfrac{31}{15} \\[5ex] = \dfrac{8}{15} * \dfrac{15}{31} \\[5ex] = \dfrac{8}{31} $
$ \cos\theta = \dfrac{15}{17} \\[5ex] \theta = \cos^{-1}\left(\dfrac{15}{17}\right) \\[5ex] \dfrac{\tan\theta}{1 + 2\tan\theta} \\[5ex] $
M sails at 35 km/hr on a bearing of 230° while N sails at 30km/hr on a bearing of 320°.
If the two canoes sailed for 2.5 hours, find, correct to one decimal place, the:
(a.) distance between M and N
(b.) bearing of M from N
$ s...t...d \\[3ex] speed * time = distance \\[3ex] \underline{\text{Canoe M}} \\[3ex] speed = 35\;km/hr \\[3ex] time = 2.5\;hr \\[3ex] distance = 35(2.5) = 87.5\;km \\[5ex] \underline{\text{Canoe N}} \\[3ex] speed = 30\;km/hr \\[3ex] time = 2.5\;hr \\[3ex] distance = 30(2.5) = 75\;km \\[3ex] $ Let us represent the information on a diagram
$ a = |MN| \\[3ex] \underline{\text{Right Triangle MNA}} \\[3ex] (a.) \\[3ex] a^2 = 75^2 + 87.5^2...\text{Pythagorean Theorem} \\[3ex] a = \sqrt{75^2 + 87.5^2} \\[3ex] a = 115.2443057 \\[3ex] |MN| = a \approx 115.2\;km ...\text{to one decimal place} \\[5ex] \tan \angle MNA = \dfrac{87.5}{75} ...\text{SOHCAHTOA} \\[5ex] \angle MNA = \tan^{-1}\left(\dfrac{87.5}{75}\right) \\[5ex] \angle MNA = 49.39870535^\circ \\[3ex] (b.) \\[3ex] \text{Bearing of M from N} \\[3ex] = 90 + 50 + \angle MNA \\[3ex] = 140 + 49.39870535 \\[3ex] = 189.3987054 \\[3ex] \approx 189.4^\circ ...\text{to one decimal place} $
Find the bearing of P from R.
Let us represent the information diagrammatically
$ \angle PRQ = \alpha \\[3ex] \underline{\text{Right Triangle PQR}} \\[3ex] \tan\alpha = \dfrac{270}{420}...\text{SOHCAHTOA} \\[5ex] \alpha = \tan^{-1}\left(\dfrac{27}{42}\right) \\[5ex] \alpha = 32.73522627^\circ \\[3ex] \text{Bearing of P from R} \\[3ex] = 180 + 58 + \alpha \\[3ex] = 238 + 32.73522627 \\[3ex] = 270.73522627^\circ \\[3ex] \approx 271^\circ ...\text{to the nearest degree} $
It's area is 216cm² and $|\overline{PQ}| : |\overline{QR}|$ is 3 : 4.
Find $|\overline{PR}|$
$ \underline{\text{Right Triangle PQR}} \\[3ex] |\overline{PQ}| : |\overline{QR}| = 3 : 4 \\[3ex] \text{Let the side length} = d \\[3ex] \text{sum of ratios} = 3 + 4 = 7 \\[3ex] |\overline{PQ}| = \dfrac{3d}{7} \\[5ex] |\overline{QR}| = \dfrac{4d}{7} \\[5ex] $ Let us represent this information on a diagram
$ |\overline{PR}|^2 = |\overline{PQ}|^2 + |\overline{QR}|^2 ...\text{Pythagorean Theorem} \\[3ex] = \left(\dfrac{3d}{7}\right)^2 + \left(\dfrac{4d}{7}\right)^2 \\[5ex] = \dfrac{9d^2}{49} + \dfrac{16d^2}{49} \\[5ex] = \dfrac{25d^2}{49} \\[5ex] |\overline{PR}| = \sqrt{\dfrac{25d^2}{49}} \\[5ex] |\overline{PR}| = \dfrac{5d}{7} \\[5ex] Area = \dfrac{1}{2} * base \;* \perp height \\[5ex] Area = \dfrac{1}{2} * \dfrac{4d}{7} * \dfrac{3d}{7} \\[5ex] Area = 216\;cm^2 ...\text{Given} \\[3ex] \implies \\[3ex] \dfrac{1}{2} * \dfrac{4d}{7} * \dfrac{3d}{7} = 216 \\[5ex] d^2 = \dfrac{216 * 2 * 7 * 7}{4 * 3} \\[5ex] d^2 = 1764 \\[3ex] d = \sqrt{1764} \\[3ex] d = 42\;cm \\[5ex] |\overline{PR}| \\[3ex] = \dfrac{5d}{7} \\[5ex] = \dfrac{5 * 42}{7} \\[5ex] = 5(6) \\[3ex] = 30\;cm $
Cyclist X travels at the rate of 5 km/hr on a bearing of 049° and Cyclist Y travels at the rate of 9 km/hr on a bearing of 319°
(a.) Illustrate the information on a diagram.
(b.) After travelling for two hours, calculate, correct to the nearest whole number, the:
(i.) distance between Cyclists X and Y
(ii.) bearing of Cyclist X from Y.
(c.) Find the average speed at which Cyclist X will get to Cyclist Y in 4 hours.
Let the distance between Cyclists X and Y = q
θ, α = angles
(a.)
$ (b.) \\[3ex] \underline{\text{Right Triangle YQX}} \\[3ex] (i.) \\[3ex] q^2 = 18^2 + 10^2 ...\text{} q = \sqrt{18^2 + 10^2} \\[3ex] q = 20.59126028 \\[3ex] q \approx 21\;km ...\text{to the nearest whole number} \\[5ex] (ii.) \\[3ex] \tan\alpha = \dfrac{18}{10} ...\text{SOHCAHTOA} \\[5ex] \alpha = \tan^{-1}(1.8) \\[3ex] \alpha = 60.9453959^\circ \\[3ex] \theta + 41 = \alpha ...\text{diagram} \\[3ex] \theta = \alpha - 41 \\[3ex] \theta = 60.9453959 - 41 \\[3ex] \theta = 19.9453959^\circ \\[3ex] (ii.) \\[3ex] \text{Bearing of X from Y} \\[3ex] = 90 + \theta \\[3ex] = 90 + 19.9453959^\circ \\[3ex] = 109.9453959 \\[3ex] \approx 110^\circ ...\text{to the nearest whole number} \\[5ex] (c.) \\[3ex] s.......t.......d \\[3ex] speed * time = distance \\[3ex] speed = \dfrac{distance}{time} \\[5ex] distance = 20.59126028\;km \\[3ex] time = 4\;hours \\[3ex] speed = \dfrac{20.59126028}{4} \\[5ex] speed = 5.14781507\;km/hr $
It then flies 700 km to town R on a bearing of 165°
(a.) Illustrate the information with a diagram.
(b.) Calculate, correct to three significant figures, the distance between P and R.
(a.)
$|PR| = q$
$ (b.) \\[3ex] q^2 = 500^2 + 700^2 - 2(500)(700) * \cos 68^\circ ...\text{Cosine Law} \\[3ex] q^2 = 250000 + 490000 - 700000(0.3746065934) \\[3ex] q^2 = 740000 - 262224.6154 \\[3ex] q^2 = 477775.3846 \\[3ex] q = \sqrt{477775.3846} \\[3ex] q = 691.2129806 \\[3ex] q \approx 691\;km...\text{to 3 significant figures} $
He observes an object on the top, P of the building at an angle of elevation of 66°.
He moves directly backwards to a new point C and observes the same object at an angle of elevation of 53°.
If $|\overline{MT}| = 50\;m$;
(a.) Illustrate the information in a diagram;
(b.) Calculate, correct to one decimal place:
(i.) the height of the building
(ii.) $|\overline{MC}|$
Let:
$|\overline{PT}| = h$
$|\overline{MC}| = k$ (a.) The diagram is:
$ (b.)(i.) \\[3ex] \underline{\triangle PMT} \\[3ex] \tan 66^\circ = \dfrac{h}{50} ...\text{SOHCAHTOA} \\[5ex] h = 50\tan 66 ...eqn.(1) \\[3ex] h = 112.3018387 \\[3ex] h \approx 112.3\;m \\[5ex] \underline{\triangle PCT} \\[3ex] \tan 53^\circ = \dfrac{h}{k + 50} ...\text{SOHCAHTOA} \\[5ex] h = (k + 50)\tan 53 ...eqn.(2) \\[3ex] h = h \implies eqn.(2) = eqn.(1) \\[3ex] (ii.) \\[3ex] (k + 50)\tan 53 = 50\tan 66 \\[3ex] k + 50 = \dfrac{50\tan 66}{\tan 53} \\[5ex] k = \dfrac{50\tan 66}{\tan 53} - 50 \\[5ex] k = 34.62550538 \\[3ex] k \approx 34.6\;m $
It then flies from point B to point C 300 km due west.
(a.) Illustrate this on a diagram.
(b.) How far west, correct to the nearest km, is the aeroplane from the starting point?
(a.)
$|AC| = k$
(b.)
To solve the distance far west of C from A:
We need to calculate the distance, $|AD| = p$
$ \underline{\triangle CAB} \\[3ex] k^2 = 300^2 + 100^2 - 2(300)(100) \cos 120^\circ ...\text{Cosine Law} \\[3ex] k^2 = 90000 + 10000 - 60000(-0.5) \\[3ex] k^2 = 100000 + 30000 \\[3ex] k = \sqrt{130000} \\[3ex] k = 360.5551275\;km \\[5ex] \dfrac{\sin \beta}{300} = \dfrac{\sin 120^\circ}{k} ...\text{Sine Law} \\[5ex] \sin\beta = \dfrac{300 \sin 120}{360.5551275} \\[5ex] \beta = \sin^{-1}\left(\dfrac{300 \sin 120}{360.5551275}\right) \\[5ex] \beta = 46.10211376^\circ \\[5ex] \alpha + \beta = 60^\circ ...\text{diagram} \\[3ex] \alpha = 60 - \beta \\[3ex] \alpha = 60 - 46.10211376 \\[3ex] \alpha = 13.89788624^\circ \\[5ex] \underline{\text{Right Triangle CDA}} \\[3ex] \cos \alpha = \dfrac{p}{k} ...\text{SOHCAHTOA} \\[5ex] p = k \cos \alpha \\[3ex] p = 360.5551275 \cos (13.89788624)^\circ \\[3ex] p = 350\;km $
An engineer on the top of the tower, 110 m high observes that the angle of elevation of the top of the building and the angle of depression of the foot of the building are 38° and 22° respectively.
Calculate, correct to one decimal place, the:
(a.) distance between the tower and the building.
(b.) height of the building.
Let us represent the information diagrammatically
$ (a.) \\[3ex] \underline{\text{Right Triangle ABE}} \\[3ex] \angle ABE + 22^\circ = 90^\circ...\text{Diagram} \\[3ex] \angle ABE = 90 - 22 \\[3ex] \angle ABE = 68^\circ \\[5ex] \text{distance between the tower and the building} = |AE| \\[3ex] \tan 68^\circ = \dfrac{|AE|}{110} ...\text{SOHCAHTOA} \\[5ex] |AE| = 110 \tan 68 \\[3ex] = 272.2595539 \\[3ex] \approx 272.3\;m \\[5ex] \cos 68^\circ = \dfrac{110}{|BE|}...\text{SOHCAHTOA} \\[5ex] |BE| = \dfrac{110}{\cos 68} \\[5ex] = 293.6413879\;m \\[5ex] \underline{\text{Right Triangle BCD}} \\[3ex] \angle BCD + 38 + 90 = 180 ...\text{sum of interior angles of a triangle} \\[3ex] \angle BCD = 180 - 38 - 90 \\[3ex] \angle BCD = 52^\circ \\[3ex] \underline{\triangle BCE} \\[3ex] \text{height of the building} = |CE| \\[3ex] \dfrac{|CE|}{\sin (38 + 22)} = \dfrac{|BE|}{\sin 52} \\[5ex] |CE| = \dfrac{293.6413879 \sin 60}{\sin 52} \\[5ex] = 322.7124761 \\[3ex] \approx 322.7\;cm $
Calculate the perimeter of the minor segment.
$\left[\text{Take } \pi = \dfrac{22}{7}\right]$
Let us represent the information diagrammatically
$ \text{circle center} = O \\[3ex] \underline{\text{Length of Minor Arc AB}} \\[3ex] = \dfrac{2\pi r\theta}{360} \\[5ex] \text{where} \\[3ex] \theta = \text{central angle} = 72^\circ \\[3ex] r = \text{radius} = 24.5\;m \\[3ex] \implies \\[3ex] = \dfrac{2\pi(24.5)(72)}{360} \\[5ex] = 2 * \dfrac{22}{7} * 24.5 * 72 * \dfrac{1}{360} \\[5ex] = 30.8\;m \\[5ex] \underline{\text{Length of Chord AB}} \\[3ex] \text{Considering } \triangle AOB \\[3ex] |AB|^2 = |AO|^2 + |BO|^2 - 2 * |AO| * |BO| * \cos \angle AOB...\text{Cosine Law} \\[3ex] |AB|^2 = 24.5^2 + 24.5^2 - 2(24.5)(24.5)\cos 72^\circ \\[3ex] = 600.25 + 600.25 - 1200.5(0.3090169944) \\[3ex] = 1200.5 - 370.9749017 \\[3ex] = 829.5250983 \\[3ex] |AB| = \sqrt{829.5250983} \\[3ex] |AB| = 28.80147736\;m \\[5ex] \text{Perimeter of Minor Segment AB} \\[3ex] = \text{Length of Minor Arc AB} + \text{Length of Chord AB} \\[3ex] = 30.8 + 28.80147736 \\[3ex] = 59.60147736\;m $
Another town, T is 8km from K on a bearing of 155°.
Calculate:
(i.) to the nearest kilometer, the distance of T from J;
(ii.) to the nearest degree, the bearing of T from J
(b.) Given that $\sin\theta = \dfrac{5}{6 + x}$ and $\tan\theta = \dfrac{5}{12}$, 0° < x < 90°, find the value of x.
$ |JT| = k \\[3ex] \alpha, \beta = \angle s \\[3ex] $ Let us represent the information on a diagram.
$ (a.) (i.) \\[3ex] \underline{\text{Right Triangle KJT}} \\[3ex] k^2 = 8^2 + 20^2 ...\text{Pythagorean Theorem} \\[3ex] k = \sqrt{64 + 400} \\[3ex] k = 21.54065923 \\[3ex] k \approx 22\;km ...\text{to the nearest kilometer} \\[5ex] \tan\alpha = \dfrac{8}{20} ...\text{SOHCAHTOA} \\[5ex] \alpha = \tan{-1}\left(\dfrac{8}{20}\right) \\[5ex] \alpha = 21.80140949^\circ \\[3ex] \alpha + \beta = 65^\circ ...\text{diagram} \\[3ex] \beta = 65 - \alpha \\[3ex] \beta = 65 - 21.80140949 \\[3ex] \beta = 43.19859051^\circ \\[5ex] (ii.) \\[3ex] \text{Bearing of T from J} \\[3ex] = 180 + \beta \\[3ex] = 180 + 43.19859051 \\[3ex] = 223.1985905 \\[3ex] \approx 223^\circ ...\text{to the nearest degree} \\[5ex] (b.) \\[3ex] \underline{\text{SOHCAHTOA}} \\[3ex] \sin\theta = \dfrac{5}{6 + x} = \dfrac{opp}{hyp} \\[5ex] \tan\theta = \dfrac{5}{12} = \dfrac{opp}{adj} \\[5ex] opp = opp = 5 \\[3ex] hyp = 6 + x \\[3ex] adj = 12 \\[5ex] hyp^2 = opp^2 + adj^2 ...\text{Pythagorean Theorem} \\[3ex] (6 + x)^2 = 5^2 + 12^2 \\[3ex] 6 + x = \sqrt{25 + 144}...0^\circ \lt x \lt 90^\circ \\[3ex] 6 + x = 13 \\[3ex] x = 13 - 6 \\[3ex] x = 7 $
In the diagram, $|\overline{PR}| = 20$ cm, $|\overline{PS}| = 24.12$ cm and Q is a point on $\overline{PR}$.
If $\angle SPQ = 34^\circ$ and $\angle PSQ = 20^\circ$, calculate, correct to one decimal place:
(a.) $|\overline{SR}|$
(b.) $|\overline{PQ}|$
(c.) Area of $\triangle PQS$
$ (a.) \\[3ex] \underline{\triangle PSR} \\[3ex] \tan \angle SPR = \dfrac{|SR|}{|PR|} ...\text{SOHCAHTOA} \\[5ex] |SR| = |PR|\tan \angle SPR \\[3ex] = 20\tan 34^\circ \\[3ex] = 13.49017034 \\[3ex] \approx 13.5\;cm \\[5ex] (b.) \\[3ex] \underline{\triangle PQS} \\[3ex] \angle PQS + \angle SPQ + \angle PSQ = 180^\circ...\text{sum of angles in a triangle} \\[3ex] \angle PQS + 34 + 20 = 180 \\[3ex] \angle PQS = 180 - 34 - 20 \\[3ex] \angle PQS = 126^\circ \\[5ex] \dfrac{|PQ|}{\sin 20} = \dfrac{24.12}{\sin 126} ...\text{Sine Law} \\[5ex] |PQ| = \dfrac{24.12\sin 20}{\sin 126} \\[5ex] = 10.19697474 \\[3ex] \approx 10.2\;cm \\[5ex] \text{Area} = \dfrac{1}{2} * |PS| * |PQ| * \sin \angle SPQ \\[5ex] = \dfrac{1}{2} * 24.12 * 10.19697474 * \sin 34 \\[5ex] = 68.7670355 \\[3ex] \approx 68.8\;cm^2 $
In the diagram, PQR is a triangle.
If $x + y = 210^\circ$, find the value of n.
$ y + \angle QPR = 180^\circ ...\text{sum of angles on a straight line} \\[3ex] \angle QPR = 180 - y \\[3ex] x = (180 - y) + n ...\text{exterior angle of a triangle is the sum of the interior opposite angles} \\[3ex] x = 180 - y + n \\[3ex] n = x - 180 + y \\[3ex] n = 210 - 180 \\[3ex] n = 30^\circ $
If the bird is 10 m vertically above the ground,
(a.) illustrate the information on a diagram
(b.) find, correct to the nearest degree, the angle of depression of the foot of the tree.
(a.) Let the angle of depression = β
$ \sin \beta = \dfrac{10}{12.8} \\[5ex] \beta = \sin^{-1}\left(\dfrac{10}{12.8}\right) \\[5ex] \beta = 51.37516713 \\[3ex] \beta \approx 51^\circ...\text{to the nearest degree} $
