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
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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.
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]
$
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
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 other
two sides.
$hyp^2 = leg^2 + leg^2$
x is any trigonometric ratio
y is any trigonometric ratio
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]
$
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 $
$ A.\;\; N\;50^\circ\;W \\[4ex] B.\;\; N\;40^\circ\;W \\[4ex] C.\;\; N\;50^\circ\;E \\[4ex] D.\;\; N\;40^\circ\;E \\[4ex] $
$ 320 + \phi = 360 ...\angle s\;\;at\;\;a\;\;point \\[3ex] \phi = 360 - 320 \\[3ex] \phi = 40^\circ \\[3ex] Compass\;\;Bearing = N\;40^\circ\;W $
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} = 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
$ (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 $
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 $
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.) 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 $
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 $
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
$ \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} $
