Trigonometry
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These are the solutions to the GCSE Mathematics 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}
$
Work out the size of angle x.
$ \underline{\text{Triangle ACD}} \\[3ex] \angle ADC + \angle ACD + \angle CAD = 180^\circ ...\text{sum of angles of a triangle} \\[4ex] \angle ADC + 90^\circ + 37^\circ = 180^\circ \\[4ex] \angle ADC = 180 - 90 - 37 \\[3ex] \angle ADC = 53^\circ \\[5ex] \underline{\text{Triangle ABD}} \\[3ex] \dfrac{\sin\angle ABD}{|AD|} = \dfrac{\sin\angle BDA}{|AB|} ...\text{Sine Law} \\[5ex] \angle BDA = \angle ADC = 53^\circ...diagram \\[4ex] \dfrac{\sin x}{4} = \dfrac{\sin 53^\circ}{9.3} \\[5ex] \sin x = \dfrac{4\sin 53^\circ}{9.3} \\[5ex] \sin x = 0.3434991441 \\[3ex] x = \sin^{-1}(0.3434991441) \\[4ex] x = 20.09020484^\circ $
In this right-angled triangle,
$ a = 16\;cm \\[3ex] a\;:\;c = 4\;:\;5 \\[3ex] $ Work out the area of the triangle.
$ a = 16\;cm \\[3ex] a\;:\;c = 4\;:\;5 \\[3ex] \dfrac{a}{c} = \dfrac{4}{5} = \dfrac{16}{c}...\text{Equivalent Fractions} \\[5ex] 4 * \color{blue}{4} = 16...\text{Numerator — Numerator Relationship} \\[3ex] 5 * \color{blue}{4} = c = 20 ...\text{Denominator — Denominator Relationship} \\[3ex] hyp^2 = leg^2 + leg^2 ...\text{Pythagorean Theorem} \\[4ex] c^2 = a^2 + b^2 \\[4ex] b^2 = c^2 - a^2 \\[4ex] = 20^2 - 16^2 \\[4ex] = 400 - 256 \\[3ex] = 144 \\[3ex] b = \sqrt{144} \\[3ex] b = 12\;cm \\[5ex] Area = \dfrac{base \;*\; \perp height}{2} \\[5ex] = \dfrac{12 * 16}{2} \\[5ex] = 96\;cm^2 $
A boat sails 35 km North from A to B.
From B the boat sails to C and then back to A.
(a.) Show that the distance the boat sails from C to A is 79 km to the nearest km.
You must show your working.
(b.) Work out the bearing of A from C.
$ Let: \\[3ex] |AB| = c = 35\;km \\[3ex] |AC| = b = ? \\[3ex] |BC| = a = 65\;km \\[3ex] (a.) \\[3ex] b^2 = a^2 + c^2 - 2ac \cos \angle B ...\text{Cosine Law} \\[4ex] = 65^2 + 35^2 - 2(65)(35) \cos 100^\circ \\[4ex] = 4225 + 1225 - 4550(-0.1736481777) \\[3ex] = 5450 + 790.0992085 \\[3ex] = 6240.099209 \\[3ex] b = \sqrt{6240.099209} \\[3ex] = 78.99429858 \\[3ex] b \approx 79\;km...\text{to the nearest km} \\[3ex] $
$ \theta = \theta ...\text{alternate interior angles are congruent} \\[3ex] \dfrac{\sin A}{a} = \dfrac{\sin B}{b} ...\text{Sine Law} \\[5ex] \dfrac{\sin\theta}{65} = \dfrac{\sin 100^\circ}{78.99429858} \\[5ex] \sin\theta = \dfrac{65\sin 100}{78.99429858} \\[5ex] \theta = \sin^{-1}\left(\dfrac{65\sin 100}{78.99429858}\right) \\[6ex] \theta = 54.12949048^\circ \\[4ex] $ (b.) The bearing of A from C (indicated by the red arrow) = 180 + θ
= 180° + 54.12949048°
= 234.1294905°
Calculate the value of y.
$ \underline{\triangle ABC} \\[3ex] \angle BAC = \angle BCA = y^\circ ...\text{base angles of isosceles triangle} \\[4ex] \angle BAC + \angle BCA + \angle ABC = 180^\circ ...\text{sum of angles of a triangle} \\[4ex] y + y + 78 = 180 \\[3ex] 2y = 180 - 78 \\[3ex] 2y = 102 \\[3ex] y = \dfrac{102}{2} \\[5ex] y = 51^\circ $
$B\hat{A}C = 90^\circ,\;\;\;B\hat{D}E = 90^\circ\;\;\;and\;\;\;C\hat{B}E = 90^\circ$.
AC = 7.7cm, BC = 11.3cm and BD = 13.1 cm.
(a.) Calculate the value of x
(b.) Hence find the length DE.
$ (a.) \\[3ex] \underline{\triangle BAC} \\[3ex] \sin x = \dfrac{opp}{hyp} = \dfrac{7.7}{11.3}...SOHCAHTOA \\[5ex] \sin x = 0.6814159292 \\[3ex] x = \sin^{-1}{(0.6814159292)} \\[3ex] x = 42.95438807 \\[3ex] x \approx 43^\circ \\[5ex] \underline{\text{Straight Line AD}} \\[3ex] \angle EBD + 90^\circ + x = 180^\circ ...\text{sum of angles on a straight line} \\[3ex] \angle EBD + 90 + 42.95438807 = 180 \\[3ex] \angle EBD = 180 - 90 - 42.95438807 \\[3ex] \angle EBD = 47.04561193^\circ \\[5ex] (b.) \\[3ex] \underline{\triangle BED} \\[3ex] \tan \angle EBD = \dfrac{opp}{adj} = \dfrac{|ED|}{|BD|} ...SOHCAHTOA \\[5ex] |ED| = |BD|\tan \angle EBD \\[3ex] = 13.1\tan 47.04561193^\circ \\[3ex] = 13.1(1.074081721) \\[3ex] = 14.07047054 \\[3ex] \approx 14.1\;cm $
D is a point on AC where BD = 7cm and DC = 5cm.
Calculate the size of $B\hat{A}D$.
You must show all your working.
We can solve this question using at least two approaches.
Use any approach you prefer.
$ \underline{\text{1st Approach: Cosine Law and Sine Law}} \\[3ex] \underline{\triangle BDC} \\[3ex] |BC|^2 = |BD|^2 + |DC|^2 - 2 * |BD| * |DC| * \cos \angle BDC ...\text{Cosine Law} \\[3ex] 11^2 = 7^2 + 5^2 - 2(7)(5) \cos \angle BDC \\[3ex] 121 = 49 + 25 - 70 \cos \angle BDC \\[3ex] 70\cos\angle BDC = 49 + 25 - 121 \\[3ex] 70\cos\angle BDC = -47 \\[3ex] \cos\angle BDC = -\dfrac{47}{70} \\[5ex] \angle BDC = \cos^{-1}\left(-\dfrac{47}{70}\right) \\[5ex] \angle BDC = 132.1774186^\circ \\[5ex] \underline{\text{Straight Line ADC}} \\[3ex] \angle ADB + \angle BDC = 180^\circ ...\text{sum of angles on a straight line} \\[3ex] \angle ADB + 132.1774186 = 180 \\[3ex] \angle ADB = 180 - 132.1774186 \\[3ex] \angle ADB = 47.8225814^\circ \\[5ex] \underline{\triangle ABD} \\[3ex] \dfrac{\sin\angle BAD}{|BD|} = \dfrac{\sin \angle ADB}{|AB|}...\text{Sine Law} \\[3ex] \dfrac{\sin\angle BAD}{7} = \dfrac{\sin \angle ADB}{13} \\[5ex] \sin\angle BAD = \dfrac{7\sin \angle ADB}{13} \\[5ex] = \dfrac{7(\sin 47.8225814)}{13} \\[5ex] = \dfrac{7(0.7410692771)}{13} \\[5ex] = 0.3990373031 \\[3ex] \angle BAD = \sin^{-1}(0.3990373031) \\[3ex] \angle BAD = 23.51800943^\circ \\[3ex] $ In the 2nd Approach, we shall use only the Cosine Law
$ \underline{\text{1st Approach: Cosine Law}} \\[3ex] Let\;\;|AD| = k \\[3ex] Let\;\;\angle BAD = A \\[3ex] \underline{\triangle ABD} \\[3ex] |BD|^2 = |AB|^2 + |AD|^2 - 2 * |AB| * |AD| * \cos \angle BAD ...\text{Cosine Law} \\[3ex] 7^2 = 13^2 + k^2 - 2(13)(k)\cos A \\[3ex] 2(13)(k)\cos A = 13^2 + k^2 - 7^2 \\[3ex] 26k\cos A = k^2 + 120 \\[3ex] \cos A = \dfrac{k^2 + 120}{26k} ...eqn.(1) \\[5ex] \underline{\triangle ABC} \\[3ex] |BC|^2 = |AB|^2 + |AC|^2 - 2 * |AB| * |AC| * \cos \angle BAD ...\text{Cosine Law} \\[3ex] 11^2 = 13^2 + (k + 5)^2 - 2(13)(k + 5) \cos A \\[3ex] 2(13)(k + 5)\cos A = 13^2 + (k + 5)^2 - 11^2 \\[3ex] 26(k + 5)\cos A = (k + 5)(k + 5) + 48 \\[3ex] (26k + 130)\cos A = k^2 + 5k + 5k + 25 + 48 \\[3ex] \cos A = \dfrac{k^2 + 10k + 73}{26k + 130} ...eqn.(2) \\[5ex] eqn.(1) = eqn.(2) \\[3ex] \implies \\[3ex] \dfrac{k^2 + 120}{26k} = \dfrac{k^2 + 10k + 73}{26k + 130} \\[5ex] (k^2 + 120)(26k + 130) = 26k(k^2 + 10k + 73) \\[3ex] 26k^3 + 130k^2 + 3120k + 15600 = 26k^3 + 260k^2 + 1898k \\[3ex] 26k^3 + 260k^2 + 1898k = 26k^3 + 130k^2 + 3120k + 15600 \\[3ex] 26k^3 - 26k^3 + 260k^2 - 130k^2 + 1898k - 3120k - 15600 = 0 \\[3ex] 130k^2 - 1222k - 15600 = 0 \\[3ex] \text{Divide both sides by 26 to simplify the equation} \\[3ex] 5k^2 - 47k - 600 = 0 \\[3ex] \text{Compare with } ax^2 + bx + c = 0 \\[3ex] a = 5 \\[3ex] b = -47 \\[3ex] c = -600 \\[3ex] k = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}...\text{Quadratic Formula} \\[5ex] k = \dfrac{-(-47) \pm \sqrt{(-47)^2 - 4(5)(-600)}}{2(5)} \\[5ex] = \dfrac{47 \pm \sqrt{2209 + 12000}}{10} \\[5ex] = \dfrac{47 \pm \sqrt{14209}}{10} \\[5ex] = \dfrac{47 \pm 119.2015101}{10} \\[5ex] \text{Discard the negative value because the side length cannot be negative} \\[3ex] \implies \\[3ex] k = \dfrac{47 + 119.2015101}{10} \\[5ex] = \dfrac{166.2015101}{10} \\[5ex] = 16.62015101\;cm \\[5ex] \text{Substitute the value of k in eqn. (1)} \\[3ex] \cos A = \dfrac{16.62015101^2 + 120}{26(16.62015101)} \\[5ex] \cos A = 0.9169346928 \\[3ex] A = \cos^{-1}(0.9169346928) \\[3ex] A = 23.51800942 \\[3ex] \therefore \angle BAD = 23.51800942^\circ $
