Trigonometry
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These are the solutions to the CSEC 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.
Formula Sheet: List of Formulae
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]
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]
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]
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}
$
(i.) Calculate the length of QR.
(ii.) The bearing of P from S is 205°
Determine the bearing of
(a.) R from S
(b.) S from P
$ (i.) \\[3ex] \underline{\triangle QRS} \\[3ex] |QR|^2 = |QS|^2 + |RS|^2 - 2 \cdot |QS| \cdot |RS| \cos \angle QSR...\text{Cosine Law} \\[4ex] = 18^2 + 16^2 - 2(18)(16) \cdot \cos 25^\circ \\[4ex] = 324 + 256 - 576(0906307787) \\[3ex] = 57.96671467\;m \\[3ex] $ (ii.) Let us update the information on the diagram
Please observe the colors.
$ \text{Bearing of P from S} = 205^\circ ...\text{red color} \\[4ex] (a.) \\[3ex] \text{Bearing of R from S} = \phi^\circ ...\text{dark blue color} \\[4ex] \phi + 25^\circ + 72^\circ = 205^\circ...diagram \\[4ex] \phi = 205 - 25 - 72 \\[3ex] \phi = 108^\circ \\[5ex] (b.) \\[3ex] \text{Bearing of S from P} = \beta^\circ ...\text{green color} \\[4ex] \alpha = 180^\circ ...\text{angle on a straight line}...\text{brown color} \\[4ex] \beta = \beta ...\text{alternate interior angles are congruent} \\[4ex] \beta + \alpha = 205^\circ ...diagram \\[3ex] \beta + 180 = 205 \\[3ex] \beta = 205 - 180 \\[3ex] \beta = 25^\circ $
(i.) Calculate the value of Angle MLP.
(ii.) The bearing of P from L is 210°.
(a.) Find the bearing of M from L.
(b.) Calculate the value of Angle NLP and hence, find the bearing of L from P.
Let:
|LP| = m
|LM| = p
|PM| = l
$\angle MLP = \hat{L}$
$ (i.) \\[3ex] \underline{\text{Cosine Law}} \\[3ex] \triangle PLM \\[3ex] l^2 = p^2 + m^2 - 2pm \cos \hat{L} \\[3ex] 180^2 = 120^2 + 150^2 - 2(120)(150) \cos \hat{L} \\[3ex] 32400 = 14400 + 22500 - 36000 \cos \hat{L} \\[3ex] 36000\cos \hat{L} = 14400 + 22500 - 32400 \\[3ex] \cos \hat{L} = \dfrac{4500}{36000} \\[5ex] \cos \hat{L} = 0.125 \\[3ex] \hat{L} = \cos^{-1}(0.125) \\[3ex] \hat{L} = 82.81924422^\circ \\[3ex] $ Let us represent the information on the diagram
Please see the colors for the angles
The bearing of M from L = α
Angle NLP = β
Bearing of L from P = τ
$ (ii.)(a.) \\[3ex] \alpha + \hat{L} = 210^\circ ...diagram \\[3ex] \alpha + 82.81924422 = 210 \\[3ex] \alpha = 210 - 82.81924422 \\[3ex] \alpha = 127.1807558^\circ \\[5ex] (b.) \\[3ex] \beta + 210^\circ = 360^\circ ...\text{angles at a point} \\[3ex] \beta = 360 - 210 \\[3ex] \beta = 150^\circ \\[5ex] \tau = \tau ...\text{interior alternate angles are congruent} \\[3ex] \beta + \tau = 180^\circ ...\text{angles on a straight line} \\[3ex] \tau = 180 - \beta \\[3ex] \tau = 180 - 150 \\[3ex] \tau = 30^\circ $
It then travels to Comptin (C) which is 310 km due east of Akron (A), as shown in the diagram below.
(i.) Indicate on the diagram the bearing 030° and the distances 90 km and 310 km
(ii.) Calculate, to the nearest km, the distance between Bellville (B) and Comptin (C)
(iii.) Calculate, to the nearest degree, the measure of $A\hat{B}C$
(iv.) Determine the bearing of Comptin (C) from Bellville (B)
(i.)
$ (ii.) \\[3ex] \underline{\triangle ABC} \\[3ex] B\hat{A}C + 30 = 90 ...perpendicular\;\;angle \\[3ex] B\hat{A}C = 90 - 30 = 60^\circ \\[3ex] |BC|^2 = |AB|^2 + |AC|^2 - 2(|AB|)(|AC|) \cos B\hat{A}C \\[3ex] |BC|^2 = 90^2 + 310^2 - 2(90)(310) * \cos 60 \\[3ex] |BC|^2 = 8100 + 96100 - 55800 * 0.5 \\[3ex] |BC|^2 = 104200 - 27900 \\[3ex] |BC|^2 = 76300 \\[3ex] |BC| = \sqrt{76300} \\[3ex] |BC| = 276.2245463 \\[3ex] |BC| \approx 276\;km \\[5ex] (iii.) \\[3ex] \underline{\triangle ABC} \\[3ex] \dfrac{\sin A\hat{B}C}{310} = \dfrac{\sin B\hat{A}C}{|BC|}...Sine\;\;Law \\[5ex] \dfrac{\sin A\hat{B}C}{310} = \dfrac{\sin 60}{276.2245463} \\[5ex] \sin A\hat{B}C = \dfrac{310 * \sin 60}{276.2245463} \\[5ex] \sin A\hat{B}C = \dfrac{310 * 0.8660254038}{276.2245463} \\[5ex] \sin A\hat{B}C = \dfrac{268.4678752}{276.2245463} \\[5ex] \sin A\hat{B}C = 0.9719189651 \\[3ex] A\hat{B}C = \sin^{-1}(0.9719189651) \\[3ex] A\hat{B}C = 76.38976123 \\[3ex] A\hat{B}C \approx 76^\circ \\[5ex] (iv.) \\[3ex] $
$ Bearing\;\;of\;\;Comptin\;\;from\;\;Bellville = \theta \\[3ex] \phi = 30^\circ...alternate\;\; \angle s\;\;are\;\;equal \\[3ex] \phi + \tau = A\hat{B}C ...diagram \\[3ex] 30 + \tau = 76.38976123 \\[3ex] \tau = 76.38976123 - 30 \\[3ex] \tau = 46.38976123 \\[3ex] \theta + \tau = 180...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \theta = 180 - \tau \\[3ex] \theta = 180 - 46.38976123 \\[3ex] \theta = 133.6102388 \\[3ex] \theta \approx 133^\circ $
The diagram below gives information on the distances and angles between the points.
(i.) Show that the value of x is 29.5°, correct to 1 decimal place.
(ii.) A vertical tower, GT, is constructed at the point G and is pivoted to the ground at the points E, F and H using pieces of wire.
The angle of elevation of the top of the tower, T, from the point F is 31°
What length of wire was used to secure Point T to Point F?
(iii.) The bearing of E from H is 228°.
Find the bearing of:
(a.) H from E
(b.) G from H
$ (i.) \\[3ex] \underline{\text{Cosine Law}} \\[3ex] |GF|^2 = |GH|^2 + |HF|^2 - 2\cdot |GH|\cdot |HF|\cos\angle GHF \\[4ex] 6.9^2 = 5.3^2 + 11^2 - 2(5.3)(11)\cos x^\circ \\[4ex] 2(5.3)(11)\cos x = 5.3^2 + 11^2 - 6.9^2 \\[4ex] 116.6\cos x = 101.48 \\[3ex] \cos x = \dfrac{101.48}{116.6} \\[5ex] \cos x = 0.8703259005 \\[3ex] x = \cos^{-1}(0.8703259005) \\[4ex] x = 29.5034665 \\[3ex] x \approx 29.5^\circ...\text{to 1 decimal place} \\[3ex] $ (ii.) Indicate the information on the diagram.
Draw the vertical line: |GT| perpendicular at G
Show the angle of elevation of T from F
The length of wire was used to secure Point T to Point F = |FT|
$ \underline{\triangle TGF} \\[3ex] \cos 31^\circ = \dfrac{|FG|}{|FT|}...SOHCAHTOA \\[5ex] \cos 31 = \dfrac{6.9}{|FT|} \\[5ex] |FT| = \dfrac{6.9}{\cos 31} \\[5ex] = \dfrac{6.9}{0.8571673007} \\[5ex] = 8.049770441\;m \\[3ex] $ (iii.) Let us indicate these bearings on the diagram
Let the bearing of H from E = α
Angle on a straight line = 180°
Let the bearing of G from H = β
$ (a.) \\[3ex] \alpha = \alpha ...\text{alternate interior angles are congruent} \\[3ex] \alpha = 228^\circ - 180^\circ ...diagram \\[4ex] \alpha = 48^\circ \\[5ex] (b.) \\[3ex] \beta = 228^\circ - 54^\circ ...diagram \\[4ex] \beta = 174^\circ $
The bearing of Q from P is 133° and the angle PQR is 56°
(i) Calculate the value of angle w
(ii) Determine the bearing of P from Q
(iii) Calculate the distance RP
$ (i) \\[3ex] 56 + w = 133 ...\text{alternate interior angles are equal} \\[3ex] w = 133 - 56 \\[3ex] w = 77^\circ \\[5ex] (ii) \\[3ex] Bearing\;\;of\;\;P\;\;from\;\;Q \\[3ex] = 180 + w + 56 \\[3ex] = 180 + 133 \\[3ex] = 313^\circ \\[5ex] (iii) \\[3ex] \underline{\text{Cosine Law}} \\[3ex] |RP|^2 = 210^2 + 290^2 - 2(210)(290) \cos 56^\circ \\[4ex] |RP|^2 = 44100 + 84100 - 121800 * 0.5591929035 \\[3ex] |RP|^2 = 128200 - 68109.69564 \\[3ex] |RP|^2 = 60090.30436 \\[3ex] |RP| = \sqrt{60090.30436} \\[3ex] |RP| = 245.133238 \\[3ex] |RP| \approx 245\;km $
FT is a vertical tower.
FSR is a straight line and RS = 150 m.
The angles of elevation of T, the top of a tower, from R and S, are 22° and 40° respectively.
F is the foot of the tower.
Calculate, giving your answer to 1 decimal place where appropriate
(i) the measure of ∠RTS
(ii) the length of ST
(iii) the height of the tower, FT.
$ \angle RST + \angle FST = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle RST + 40 = 180 \\[3ex] \angle RST = 180 - 40 \\[3ex] \angle RST = 140^\circ \\[5ex] (i) \\[3ex] \underline{\triangle TRS} \\[3ex] \angle RTS + \angle RST + \angle TRS = 180^\circ ...sum\;\;of\;\;\angle s\;\;in\;\;a\;\;\triangle \\[3ex] \angle RTS + 140 + 22 = 180 \\[3ex] \angle RTS + 162 = 180 \\[3ex] \angle RTS = 180 - 162 \\[3ex] \angle RTS = 18^\circ \\[5ex] (ii) \\[3ex] \dfrac{|ST|}{\sin \angle TRS} = \dfrac{|RS|}{\sin \angle RTS}...Sine\;\;Law \\[5ex] \dfrac{|ST|}{\sin 22^\circ} = \dfrac{150}{\sin 18^\circ} \\[5ex] |ST| = \dfrac{150\sin 22^\circ}{\sin 18^\circ} \\[5ex] |ST| = \dfrac{150(0.3746065934)}{0.3090169944} \\[5ex] |ST| = 181.8378602 \\[3ex] |ST| \approx 181.8\;m \\[5ex] (iii) \\[3ex] \underline{\triangle TSF} \\[3ex] \sin \angle TSF = \dfrac{|FT|}{ST|}...SOHCAHTOA \\[5ex] \sin 40^\circ = \dfrac{|FT|}{181.8378602} \\[5ex] |FT| = 181.8378602 \sin 40 \\[3ex] |FT| = 181.8378602(0.6427876097) \\[3ex] |FT| = 116.8831235 \\[3ex] |FT| \approx 116.9\;m $
The bearing of E is 302°.
A ship is anchored at F, some distance away from G, on a bearing of 228°.
Angle EFG = 52°
This information is shown on the diagram below.
(a.) On the diagram above, insert the angle 228°, the bearing of F from G
(b.) Determine the value of Angle FEG.
(c.) Calculate GF, the distance the ship is from Port G.
(d.) Indicate the point H on the line EF, such that GH is the SHORTEST distance from G to EF.
(e.) Determine the distance GH.
(a.)
The red color in the diagram indicates 228°
Angle on a straight line = 180°
180° + 48° = 228°
(b.) Let us determine the value of Angle FGE first
Check out the blue color in the diagram
The bearing of E is 302°.
Now, we can determine the value of Angle FEG
$ \underline{\triangle EFG} \\[3ex] \angle FEG + \angle FGE + \angle EFG = 180^\circ ...\text{sum of angles of a triangle} \\[4ex] \angle FEG + 74 + 52 = 180 \\[3ex] \angle FEG = 180 - 74 - 52 \\[3ex] \angle FEG = 54^\circ \\[4ex] (c.) \\[3ex] \dfrac{|GF|}{\sin \angle FEG} = \dfrac{|EG|}{\sin \angle EFG} ...\text{Sine Law} \\[5ex] \dfrac{|GF|}{\sin 54^\circ} = \dfrac{245}{\sin 52^\circ} \\[6ex] |GF| = \dfrac{245\sin 54^\circ}{\sin 52^\circ} \\[6ex] = \dfrac{245(0.8090169944)}{0.7880107536} \\[5ex] = 251.531039\;km \\[3ex] $ (d.) The shortest distance between a point and a line is always along the line perpendicular to it.
Therefore, the point H on the line EF, such that GH is the SHORTEST distance from G to EF is the perpendicular distance from the point G to the line EF at H
$ (e.) \\[3ex] \dfrac{|GH|}{\sin \angle HFG} = \dfrac{|GF|}{\sin \angle GHF} ...\text{Sine Law} \\[5ex] \dfrac{|GH|}{\sin 52^\circ} = \dfrac{251.531039}{\sin 90^\circ} \\[6ex] |GF| = \dfrac{251.531039\sin 52^\circ}{\sin 90^\circ} \\[6ex] = \dfrac{251.531039(0.7880107536)}{1} \\[5ex] = 198.2091636\;km $
Q is 5.4 km from H while S is 3.5 km from H
(i) On the diagram below, which shows the sketch of this information, insert the value of the marked angle, QHS
(ii) Calculate QS, the distance between the two buoys.
(iii) Calculate the bearing of S from Q
$ (i) \\[3ex] \angle QHS = 311 - 185 = 216^\circ \\[3ex] $
$ (ii) \\[3ex] Let\;\; QS = h \\[3ex] h^2 = 5.4^2 + 3.5^2 - 2(5.4)(3.5) \cos 126^\circ...Cosine\;\;Law \\[3ex] h^2 = 29.16 + 12.25 - 37.8 * -0.5877852523 \\[3ex] h^2 = 41.41 + 22.21828254 \\[3ex] h^2 = 63.62828254 \\[3ex] h = \sqrt{63.62828254} \\[3ex] h = 7.976733826 \\[3ex] QS \approx 8.0\;km \\[3ex] $
$ k + 126 + 185 = 360...\angle s\;\;at\;\;a\;\;point \\[3ex] k + 311 = 360 \\[3ex] k = 360 - 311 \\[3ex] k = 49^\circ \\[3ex] p = p ...alternate\;\; \angle s \\[3ex] p + k = 90 ...right\;\; \angle \\[3ex] p = 90 - k \\[3ex] p = 90 - 49 \\[3ex] p = 41^\circ \\[3ex] Considering\;\; \triangle QHS \\[3ex] \dfrac{\sin \angle SQH}{3.5} = \dfrac{\sin 126}{h}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle SQH}{3.5} = \dfrac{0.8090169944}{7.976733826} \\[5ex] \sin \angle SQH = \dfrac{3.5 * 0.8090169944}{7.976733826} \\[5ex] \sin \angle SQH = \dfrac{2.83155948}{7.976733826} \\[5ex] \sin \angle SQH = 0.3549773055 \\[3ex] \angle SQH = \sin^{-1}(0.3549773055) \\[3ex] \angle SQH = 20.79205349^\circ \\[5ex] (iii) \\[3ex] Bearing\;\;of\;\; S\;\;from\;\;Q \\[3ex] = 90 + p + \angle SQH \\[3ex] = 90 + 41 + 20.79205349 \\[3ex] = 151.7920535 \\[3ex] \approx 152^\circ $
The angles of depression are 20° and 12° respectively.
The ships are 110 m apart.
(i) Complete the diagram below by inserting the angles of depression and the distance between the ships.
(ii) Determine, to the nearest metre,
(a) the distance, TS2, between the top of the lighthouse and Ship 2
(b) the height of the lighthouse, TB.
(i)
The angles of depression and the distance between the ships are shown in the diagram below:
$ (ii) \\[3ex] \angle TS_1B = 20^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \angle TS_1S_2 + \angle TS_1B = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \angle TS_1S_2 + 20 = 180 \\[3ex] \angle TS_1S_2 = 180 - 20 \\[3ex] \angle TS_1S_2 = 160^\circ \\[3ex] \underline{\triangle TS_2S_1} \\[5ex] (a) \\[3ex] \dfrac{|TS_2|}{\sin\angle TS_1S_2} = \dfrac{|S_1S_2|}{\sin\angle S_1TS_2}...Sine\;\;Law \\[5ex] \dfrac{|TS_2|}{\sin 160^\circ} = \dfrac{110}{\sin 8^\circ} \\[5ex] |TS_2| = \dfrac{110\sin 160^\circ}{\sin 8^\circ} \\[5ex] |TS_2| = \dfrac{110(0.3420201433)}{0.139173101} \\[5ex] |TS_2| = 270.3267766\;m \\[3ex] |TS_2| \approx 270\;m \\[5ex] (b) \\[3ex] \angle TS_2B = 12^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \underline{\triangle TS_2B} \\[3ex] \sin \angle TS_2B = \dfrac{|TB|}{|TS_2|}...SOHCAHTOA \\[5ex] \sin 12^\circ = \dfrac{|TB|}{270.3267766} \\[5ex] |TB| = 270.3267766\sin 12 \\[3ex] |TB| = 270.3267766(0.2079116908) \\[3ex] |TB| = 56.2040972 \\[3ex] |TB| \approx 56\;m $
The boat then changes course and travels for a distance of 8 km to point C on a bearing of 060°
(a.) Illustrate the above diagram in a clearly labelled diagram
The diagram should show the
(i.) north direction
(ii.) bearings 135° and 060°
(iii) distances 8 km and 15 km
(b.) Calculate
(i.) the distance AC
(ii.) $\angle BCA$
(iii.) the bearing of A from C
(a.)
$ (b.) \\[3ex] \theta = 45^\circ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \theta + \phi = 90...complementary\;\;\angle s \\[3ex] \phi = 90 - \theta \\[3ex] \phi = 90 - 45 \\[3ex] \phi = 45^\circ \\[3ex] \angle ABC = B = \phi + 60...diagram \\[3ex] B = 45 + 60 \\[3ex] B = 105^\circ \\[5ex] (i.) \\[3ex] Distance\;\; AC = b \\[3ex] b^2 = a^2 + c^2 - 2ac \cos B...Cosine\;\;Law \\[3ex] b^2 = 8^2 + 15^2 - 2(8)(15) * \cos 105 \\[3ex] b^2 = 64 + 225 - 240 * - 0.2588190451 \\[3ex] b^2 = 289 + 62.11657082 \\[3ex] b^2 = 351.1165708 \\[3ex] b = \sqrt{351.1165708} \\[3ex] b = 18.73810478 \\[3ex] b \approx 19\;km \\[5ex] (ii.) \\[3ex] \angle BCA = \angle C \\[3ex] \dfrac{\sin C}{c} = \dfrac{\sin B}{b} \\[5ex] \dfrac{\sin C}{15} = \dfrac{\sin 105}{18.73810478} \\[5ex] \sin C = \dfrac{15\sin 105}{18.73810478} \\[5ex] \sin C = \dfrac{15(0.9659258263)}{18.73810478} \\[5ex] \sin C = \dfrac{14.48888739}{18.73810478} \\[5ex] \sin C = 0.773231208 \\[3ex] C = \sin^{-1}(0.773231208) \\[3ex] C = 50.64494187 \\[3ex] C \approx 51^\circ \\[3ex] $
$ (iii) \\[3ex] \tau = 60^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;A\;\;from\;\;C \\[3ex] = 180 + \tau + \angle BCA \\[3ex] = 180 + 60 + 50.64494187 \\[3ex] = 290.6449419 \\[3ex] \approx 291^\circ $
Show clearly the north line in your diagram
Town F is 50 km east of town G
Town H is on a bearing of 040° from town F
The distance from F to H is 65 km
(ii) Calculate, to the nearest kilometre, the actual distance GH
(iii) Calculate, to the nearest degree, the bearing of H from G
(i)
$ (ii) \\[3ex] \angle GFH = 90 + 40 = 130^\circ \\[3ex] |GH|^2 = |GF|^2 + |FH|^2 - 2 * |GF| * |FH| * \cos \angle GFH...Cosine\;\;Law \\[3ex] |GH|^2 = 50^2 + 65^2 - 2(50)(65) * \cos 130 \\[3ex] |GH|^2 = 2500 + 4225 - 6500 * -0.6427876097 \\[3ex] |GH|^2 = 6725 + 4178.119463 \\[3ex] |GH|^2 = 10903.11946 \\[3ex] |GH| = \sqrt{10903.11946} \\[3ex] |GH| = 104.4180035 \\[3ex] |GH| \approx 104\;km \\[5ex] (iii) \\[3ex] \dfrac{\sin \angle HGF}{|FH|} = \dfrac{\sin \angle GFH}{|GH|}...Sine\;\;Law \\[5ex] \dfrac{\sin \phi}{65} = \dfrac{\sin 130}{104.4180035} \\[5ex] \sin \phi = \dfrac{65 \sin 130}{104.4180035} \\[5ex] \sin \phi = \dfrac{65 * 0.7660444431}{104.4180035} \\[5ex] \sin \phi = \dfrac{49.7928888}{104.4180035} \\[5ex] \sin \phi = 0.4768611459 \\[3ex] \phi = \sin^{-1}(0.4768611459) \\[3ex] \phi = 28.48059838 \\[3ex] \theta + \phi = 90 ...right\;\;\angle \\[3ex] \theta = 90 - \phi \\[3ex] \theta = 90 - 28.48059838 \\[3ex] \theta = 61.51940162 \\[3ex] Bearing\;\;of\;\;H\;\;from\;\;G = \theta \approx 62^\circ $
Q is 175 km from H while R is 242 km from H
(i) Complete the diagram above to show the information given.
(ii) Calculate QR, the distance between the two ships, to the nearest km.
(iii) Calculate how far due south is Ship R of the harbour, H
(i)
$ (ii) \\[3ex] |QR|^2 = |QH|^2 + |RH|^2 - 2 * |QH| * |RH| * \cos \angle QHR...Cosine\;\;Law \\[3ex] |QR|^2 = 175^2 + 242^2 - 2(175)(242) * \cos 82 \\[3ex] |QR|^2 = 30625 + 58564 - 84700 * 0.139173101 \\[3ex] |QR|^2 = 89189 - 11787.96165 \\[3ex] |QR|^2 = 77401.03835 \\[3ex] |QR| = \sqrt{77401.03835} \\[3ex] |QR| = 278.210421 \\[3ex] |QR| \approx 278\;km \\[5ex] $
How far due south is Ship R of the harbour, H = |PH|
$ 69 + 82 + \theta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] 151 + \theta = 180 \\[3ex] \theta = 180 - 151 \\[3ex] \theta = 29^\circ \\[3ex] \underline{\triangle HPR} \\[3ex] \cos \theta = \dfrac{|PH|}{|RH|}...SOHCAHTOA \\[3ex] \cos 29 = \dfrac{|PH|}{242} \\[5ex] 242 \cos 29 = |PH| \\[3ex] |PH| = 242 * \cos 29 \\[3ex] |PH| = 242(0.8746197071) \\[3ex] |PH| = 211.6579691 \\[3ex] |PH| \approx 212\;km $
Ship T is 180 kilometres from L on a bearing of 148°
This information is illustrated in the diagram below.
(i) Complete the diagram above by inserting the value of angle RLT
(ii) Calculate RT, the distance between the two ships.
(iii) Determine the bearing of T from R.
The information is represented on the diagram:
$ (i) \\[3ex] \angle RLT + 65^\circ = 148^\circ...bearing\;\;of\;\;T\;\;from\;\;L \\[3ex] \angle RLT = 148 - 65 \\[3ex] \angle RLT = 83^\circ \\[5ex] (ii) \\[3ex] |RT|^2 = |LT|^2 + |LR|^2 - 2 * |LT| * |LR| * \cos \angle RLT...Cosine\;\;Law \\[3ex] |RT|^2 = 180^2 + 250^2 - 2(180)(250) * \cos 83^\circ \\[3ex] |RT|^2 = 32400 + 62500 - (90000 * 0.1218693434) \\[3ex] |RT|^2 = 94900 - 10968.24091 \\[3ex] |RT|^2 = 83931.75909 \\[3ex] |RT| = \sqrt{83931.75909} \\[3ex] |RT| = 289.7097843 \\[3ex] |RT| \approx 290\;km \\[3ex] $
$ \dfrac{\sin \angle LRT}{|LT|} = \dfrac{\sin \angle RLT}{|RT|} ...Sine\;\;Law \\[5ex] \dfrac{\sin \theta}{180} = \dfrac{\sin 83}{289.7097843} \\[5ex] \sin \theta = \dfrac{180 * \sin 83}{289.7097843} \\[5ex] \sin \theta = \dfrac{178.6583073}{289.7097843} \\[5ex] \sin \theta = 0.6166802676 \\[3ex] \theta = \sin^{-1}(0.6166802676) \\[3ex] \theta = 38.07411323^\circ \\[3ex] \psi + \theta = 65^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] \psi = 65 - \theta \\[3ex] \psi = 65 - 38.07411323^\circ \\[3ex] \psi = 26.92588677^\circ \\[5ex] (iii) \\[3ex] Bearing\;\;of\;\;T\;\;from\;\;R \\[3ex] = 180 + \psi \\[3ex] = 180 + 26.92588677^\circ \\[3ex] = 206.9258868^\circ $
The distance BF = 32 km, FG = 55 km, ∠BFG is 103° and F is on a bearing of 042° from B.
(i) Determine the bearing of B from F.
(ii) Calculate the distance BG, giving your answer to one decimal place.
(iii) Calculate, to the nearest degree, the bearing of G from B.
(i)
$ \phi = 42^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;B\;\;from\;\;F\;\;(blue\;\;color) \\[3ex] = 180 + \phi \\[3ex] = 180 + 42 \\[3ex] = 222^\circ \\[3ex] (ii) \\[3ex] |BG|^2 = |BF|^2 + |FG|^2 - 2 * |BF| * |FG| * \cos \angle BFG ...Cosine\;\;Law \\[3ex] |BG|^2 = 32^2 + 55^2 - 2(32)(55) * \cos 103^\circ \\[3ex] |BG|^2 = 1024 + 3025 - 3520(-0.2249510543) \\[3ex] |BG|^2 = 4049 + 791.8277113 \\[3ex] |BG|^2 = 4840.827711 \\[3ex] |BG| = \sqrt{4840.827711} \\[3ex] |BG| = 69.57605703 \\[3ex] |BG| \approx 70\;km \\[3ex] $
$ \dfrac{\sin \angle FBG}{|FG|} = \dfrac{\sin \angle BFG}{|BG|} ...Sine\;\;Law \\[5ex] \dfrac{\sin \angle FBG}{55} = \dfrac{\sin 103^\circ}{69.57605703} \\[5ex] \sin \angle FBG = \dfrac{55\sin 103^\circ}{69.57605703} \\[5ex] \sin \angle FBG = \dfrac{55(0.9743700648)}{69.57605703} \\[5ex] \sin \angle FBG = 0.7702413136 \\[3ex] \angle FBG = \sin^{-1}(0.7702413136) \\[3ex] \angle FBG = 50.37556355^\circ \\[3ex] Bearing\;\;of\;\;G\;\;from\;\;B\;\;(red\;\;color) \\[3ex] = 42 + \angle FBG \\[3ex] = 42 + 50.37556355 \\[3ex] = 92.37556355 \\[3ex] \approx 92^\circ $
The ship then changes course to sail to Port C, 72 km away, on a bearing of 105°.
(i) On the diagram below, not drawn to scale, label the known distances travelled and the known angles.
(ii) Determine the measure of ∠ABC.
(iii) Calculate, to the nearest km, the distance AC.
(iv) Show that the bearing of A from C, to the nearest degree, is 260°.
(i) The labelled diagram is:
(ii)
Let us indicate more labels to help us with the calculations.
$ \phi = 44^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] 105 + \theta = 180^\circ ...\angle s\;\;on\;\;a\;\;straight\;\;line \\[3ex] \theta = 180 - 105 \\[3ex] \theta = 75^\circ \\[5ex] (ii) \\[3ex] \angle ABC = \phi + \theta ...diagram \\[3ex] \angle ABC = 44 + 75 \\[3ex] \angle ABC = 119^\circ \\[5ex] (iii) \\[3ex] |AC|^2 = |AB|^2 + |BC|^2 - 2 * |AB| * |BC| * \cos \angle ABC...Cosine\;\;Law \\[3ex] |AC|^2 = 52^2 + 72^2 - 2(52)(72)(\cos 119^\circ) \\[3ex] |AC|^2 = 2704 + 5184 - 7488(-0.4848096202) \\[3ex] |AC|^2 = 7888 + 3630.254436 \\[3ex] |AC|^2 = 11518.25444 \\[3ex] |AC| = \sqrt{11518.25444} \\[3ex] |AC| = 107.3231309 \\[3ex] |AC| \approx 107\;km \\[5ex] (iv) \\[3ex] \dfrac{\sin \angle ACB}{|AB|} = \dfrac{\sin \angle ABC}{|AC|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle ACB}{52} = \dfrac{\sin 119^\circ}{107.3231309} \\[5ex] \sin \angle ACB = \dfrac{52\sin 119^\circ}{107.3231309} \\[5ex] \sin \angle ACB = \dfrac{52(0.8746197071)}{107.3231309} \\[5ex] \sin \angle ACB = 0.423769083 \\[3ex] \angle ACB = \sin^{-1}(0.423769083) \\[3ex] \angle ACB = 25.07277523^\circ \\[3ex] \beta = \theta = 75^\circ ...alternate\;\;\angle s\;\;are\;\;equal \\[3ex] Bearing\;\;of\;\;A\;\;from\;\;C...indicated\;\;by\;\;\color{purple}{purple\;\;color} \\[3ex] = 360 - (\beta + \angle ACB) ...\angle s\;\;at\;\;a\;\;point \\[3ex] = 360 - (75 + 25.07277523) \\[3ex] = 360 - 100.0727752 \\[3ex] = 259.9272248 \\[3ex] \approx 260^\circ $
LR = 18 km, LN = 12 km and MN = 10 km.
Angle RLN = 25° and angle LMN = 88°
(i) Calculate angle MLN.
(ii) Calculate the distance NR.
(iii) Determine the bearing of Town R from Town L.
$ (i) \\[3ex] \underline{\triangle LMN} \\[3ex] \dfrac{\sin \angle MLN}{|MN|} = \dfrac{\sin \angle LMN}{|LN|}...Sine\;\;Law \\[5ex] \dfrac{\sin \angle MLN}{10} = \dfrac{\sin 88^\circ}{12} \\[5ex] \sin \angle MLN = \dfrac{10 * \sin 88^\circ}{12} \\[5ex] \sin \angle MLN = \dfrac{10(0.999390827)}{12} \\[5ex] \sin \angle MLN = 0.8328256892 \\[3ex] \angle MLN = \sin^{-1}(0.8328256892) \\[3ex] \angle MLN = 56.39010829 \\[3ex] \angle MLN \approx 56^\circ \\[5ex] (ii) \\[3ex] \underline{\triangle LNR} \\[3ex] |NR|^2 = |LN|^2 + |LR|^2 - 2 * |LN| * |LR| * \cos \angle RLN...Cosine\;\;Law \\[3ex] |NR|^2 = 12^2 + 18^2 - 2(12)(18) * \cos 25^\circ \\[3ex] |NR|^2 = 144 + 324 - 432 * 0.906307787 \\[3ex] |NR|^2 = 468 - 391.524964 \\[3ex] |NR|^2 = 76.475036 \\[3ex] |NR| = \sqrt{76.475036} \\[3ex] |NR| = 8.745000629 \\[3ex] |NR| \approx 9\;km \\[5ex] (iii) \\[3ex] Bearing\;\;of\;\;R\;\;from\;\;L \\[3ex] = 50 + \angle MLN + 25...diagram \\[3ex] = 50 + 56.39010829 + 25 \\[3ex] = 131.3901083 \\[3ex] \approx 131^\circ $
