Trigonometry
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The ACT Mathematics test is a timed exam...60 questions in 60 minutes
This implies that you have to solve each question in one minute.
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the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.
These are the solutions to the ACT past questions on the topics in Trigonometry.
When applicable, the TI-84 Plus CE calculator (also applicable to TI-84 Plus calculator) solutions are provided
for some questions.
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Please NOTE: For applicable questions (questions that require the angle in degrees), if you intend to
use the TI-84 Family:
Please make sure you set the MODE to DEGREE.
It is RADIAN by default. So, it needs to be set to DEGREE.
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]
$
Theorems
(1.) The sum of the interior angles of a triangle is 180°
(2.) The sum of angles on a straight line is 180°
(3.) The exterior angle of a triangle is the sum of the two interior opposite angles.
(4.)
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}
$
What is the measure of $\angle ACD$?
$ F.\;\; 95^\circ \\[3ex] G.\;\; 125^\circ \\[3ex] H.\;\; 130^\circ \\[3ex] J.\;\; 140^\circ \\[3ex] K.\;\; 145^\circ \\[3ex] $
Average implies Arithmetic Mean
$ \angle ACD = \angle BAC + \angle ABC ...exterior\;\;\angle\;\;of\;\;a\;\;\triangle = sum\;\;of\;\;the\;\;two\;\;interior\;\;opposite\;\;\angle s \\[3ex] \angle ACD = 35 + 95 \\[3ex] = 130^\circ $
What is $\sin \theta$?
$ A.\;\; \dfrac{25}{39} \\[5ex] B.\;\; \dfrac{5}{8} \\[5ex] C.\;\; \dfrac{\sqrt{39}}{8} \\[5ex] D.\;\; 1 - \dfrac{5\sqrt{39}}{39} \\[5ex] E.\;\; \sqrt{1 - \left(\dfrac{5}{\sqrt{39}}\right)^2} \\[7ex] $
$ \underline{SOHCAHTOA} \\[3ex] \tan \theta = \dfrac{5}{\sqrt{39}} = \dfrac{opp}{adj} \\[5ex] \underline{Pythagorean\;\;Theorem} \\[3ex] hyp^2 = opp^2 + adj^2 \\[3ex] hyp^2 = 5^2 + \left(\sqrt{39}\right)^2 \\[4ex] hyp^2 = 25 + 39 \\[3ex] hyp^2 = 64 \\[3ex] hyp = \sqrt{64} \\[3ex] hyp = 8\;units \\[3ex] \sin\theta = \dfrac{opp}{hyp}...SOHCAHTOA \\[5ex] \sin\theta = \dfrac{5}{8} $
What is the measure of $\angle BCD$?
$ A.\;\; 65^\circ \\[4ex] B.\;\; 85^\circ \\[4ex] C.\;\; 95^\circ \\[4ex] D.\;\; 105^\circ \\[4ex] E.\;\; 115^\circ \\[4ex] $
$ \angle BCD = 65^\circ + 30^\circ ...exterior\;\angle \;\;of\;\;a\;\;\triangle = sum\;\;of\;\;two\;\;interior\;\;opposite\;\;\angle s \\[4ex] \angle BCD = 95^\circ $
Which of the following expressions gives the value of x in terms of b?
$ A.\;\; b \\[3ex] B.\;\; \dfrac{21}{5}b \\[5ex] C.\;\; 5b \\[3ex] D.\;\; 9b \\[3ex] E.\;\; \dfrac{35}{3}b \\[5ex] $
$ \dfrac{x}{7b} = \dfrac{3a}{5a}...\text{ratio of corresponding sides of similar}\;\triangle s \\[5ex] x = \dfrac{3a \cdot 7b}{5a} \\[5ex] x = \dfrac{21b}{5} $
The perimeter of the triangle is 32 centimeters.
What is the length, in centimeters, of the altitude that splits the triangle into 2 congruent right triangles?
$ A.\;\; \sqrt{44} \\[3ex] B.\;\; 6 \\[3ex] C.\;\; 7 \\[3ex] D.\;\; 8 \\[3ex] E.\;\; 10 \\[3ex] $
Let the perpendicular altitude = h
$ \underline{2\;\;Right\;\;\triangle s} \\[3ex] Perimeter = 32\;cm \\[3ex] Length\;\;of\;\;base = 32 - (10 + 10) \\[3ex] = 32 - 20 \\[3ex] = 12\;cm \\[3ex] Base\;\;of\;\;each\;\;Right\;\;\triangle = \dfrac{12}{2} = 6\;cm \\[5ex] ...\text{altitude splits the triangle into 2 congruent right triangles} \\[3ex] \text{Using any of the Right Triangles} \\[3ex] 10^2 = h^2 + 6^2...Pythagorean\;\;Theorem \\[4ex] h^2 = 10^2 - 6^2 \\[4ex] h^2 = 100 - 36 \\[3ex] h^2 = 64 \\[3ex] h = \sqrt{64} \\[3ex] h = 8\;cm $
Their positions are at the vertices of a triangle.
Which of the following expressions gives the degree measure of the angle of the triangle at the vertex where Olivia is standing?
$ F.\;\; \cos^{-1}\left(\dfrac{40}{20}\right) \\[5ex] G.\;\; \sin^{-1}\left(\dfrac{40}{20}\right) \\[5ex] H.\;\; \sin^{-1}\left(\dfrac{20}{40}\right) \\[5ex] J.\;\; \tan^{-1}\left(\dfrac{40}{20}\right) \\[5ex] K.\;\; \tan^{-1}\left(\dfrac{20}{40}\right) \\[5ex] $
Let the degree measure of the angle of the triangle at the vertex where Olivia is standing be $\theta$
Let us represent the information using a diagram
$ SOH:CAH:TOA \\[3ex] \tan \theta = \dfrac{40}{20} \\[5ex] \theta = \tan^{-1}\left(\dfrac{40}{20}\right) $
The point of contact of the wire and the pole is 8 feet above the ground.
What angle does the wire make with the level ground?
$ A.\;\; \cos^{-1}\left(\dfrac{8}{12}\right) \\[6ex] B.\;\; \csc^{-1}\left(\dfrac{8}{12}\right) \\[6ex] C.\;\; \sec^{-1}\left(\dfrac{8}{12}\right) \\[6ex] D.\;\; \sin^{-1}\left(\dfrac{8}{12}\right) \\[6ex] E.\;\; \tan^{-1}\left(\dfrac{8}{12}\right) \\[6ex] $
Let the angle that the wire makes with the level ground = θ
$ \sin\theta = \dfrac{opp}{hyp} ...SOHCAHTOA \\[5ex] \sin\theta = \dfrac{8}{12} \\[5ex] \theta = \sin^{-1}\left(\dfrac{8}{12}\right) $
The tables below give information about the balloon, the equipment, and the tours offered.
| Hot-air-balloon information | |
|
Volume of balloon Maximum capacity of basket Weight of balloon Weight of basket Weight of burner |
$80,000$ cubic feet $8$ people $200$ pounds $150$ pounds $50$ pounds |
| Tour information | |||
| Tour | Ticket price | Duration, in minutes | Maximum altitude, in feet |
|
A B C |
$\$100$ $\$125$ $\$200$ |
$45$ $60$ $90$ |
$500$ $600$ $1,000$ |
Jarrod is looking up at a hot-air balloon.
The balloon is currently at the maximum altitude duirng Tour C
The angle of elevation from the horizon is $37^\circ$, as shown in the figure below.
Which of the following expressions is closest to the distance, $d$ feet, from Jarrod to the basket?
$ F.\;\; \dfrac{1,000}{\sin 37^\circ} \\[5ex] G.\;\; \dfrac{1,000}{\cos 37^\circ} \\[5ex] H.\;\; 1,000 \sin 37^\circ \\[3ex] J.\;\; 1,000 \cos 37^\circ \\[3ex] K.\;\; 1,000 \tan 37^\circ \\[3ex] $
$ SOH:CAH:TOA \\[3ex] \sin 37^\circ = \dfrac{opp}{hyp} = \dfrac{1000}{d} \\[6ex] d * \sin 37 = 1000 \\[3ex] d = \dfrac{1,000}{\sin 37^\circ} $
From a point 200 feet along level ground from the base of the tower, the angle of elevation to the top of the tower is 50°.
Which of the following values is closest to the height, in feet, of the radio tower?
(Note: $\sin 50^\circ \approx 0.77, \;\;\cos 50^\circ \approx 0.64, \;\;\tan 50^\circ \approx 1.20$)
$ F.\;\; 150 \\[3ex] G.\;\; 170 \\[3ex] H.\;\; 240 \\[3ex] J.\;\; 260 \\[3ex] K.\;\; 310 \\[3ex] $
Let us represent the information on a diagram
$ \underline{SOHCAHTOA} \\[3ex] \tan 50^\circ = \dfrac{opp}{adj} = \dfrac{height}{200} \\[5ex] height = 200\tan 50^\circ \\[3ex] \approx 200(1.20) \\[3ex] \approx 240\;feet $
$ F.\;\; \tan(0) \\[3ex] G.\;\; \tan(\pi) \\[3ex] H.\;\; \dfrac{1}{\sin\left(-\dfrac{\pi}{2}\right)} \\[7ex] J.\;\; \dfrac{1}{\cos(0)} \\[5ex] K.\;\; \dfrac{1}{\sin(0)} \\[5ex] $
When the argument (the angular measures) has π, then it is necessary to use the angle measure in RADIANS.
Alternatively, you may convert each angular measure in radians to degrees, and use the angle measure in DEGREES.
With your knowledge of Unit Circle (though we shall confirm with the calculator), we know that $\sin 0 = 0$
Any term that has only zero as the denominator is undefined because you cannot divide by zero.
Division by zero is undefined.
Therefore, $\dfrac{1}{\sin(0)} = \dfrac{1}{0} = undefined$
The rest of the options have values. Please see the Calculator Solution.
What is the straight-line distance, in miles, from the bakery to Juanita's home?
$ A.\;\; 0.1 \\[3ex] B.\;\; 0.2 \\[3ex] C.\;\; 0.3 \\[3ex] D.\;\; 0.5 \\[3ex] E.\;\; 0.7 \\[3ex] $
Let us represent this information in a diagram
$ Straight-line\;\;distance = hyp \\[3ex] hyp^2 = 0.3^2 + 0.4^2 ... Pythagorean\:\: Theorem \\[3ex] hyp^2 = 0.09 + 0.16 \\[3ex] hyp^2 = 0.25 \\[3ex] hyp = \sqrt{0.25} \\[3ex] hyp = 0.5\;miles $
Olivia is lying on her stomach on the diving board looking at Josh.
The horizontal and vertical distances, in meters, between Josh and Olivia are given in the diagram below.
What is the measure of the angle of elevation, $\theta$, of Josh's line of sight?
(Note: Drawing is NOT to scale.)
$ F.\;\; Arc\;sin\left(\dfrac{3}{8}\right) \\[5ex] G.\;\; Arc\;cos\left(\dfrac{3}{8}\right) \\[5ex] H.\;\; Arc\;tan\left(\dfrac{3}{8}\right) \\[5ex] J.\;\; Arc\;cot\left(\dfrac{3}{8}\right) \\[5ex] K.\;\; Arc\;csc\left(\dfrac{3}{8}\right) \\[5ex] $
$ SOHCAHTOA \\[3ex] \tan \theta = \dfrac{3}{8} \\[5ex] \theta = \tan^{-1}\left(\dfrac{3}{8}\right) \\[5ex] \theta = Arctan\left(\dfrac{3}{8}\right) $
The solution of which of the following equations gives the length q in meters?
(Note: For a triangle with sides of length a, b, and c that are opposite angles $\angle A$, $\angle B$, and $\angle C$, respectively, $\dfrac{\sin \angle A}{a} = \dfrac{\sin \angle B}{b} = \dfrac{\sin \angle C}{c}$ and $c^2 = a^2 + b^2 - 2ab \cos \angle C$.)
$ F.\;\; \dfrac{\sin q}{120} = \dfrac{8}{10} \\[5ex] G.\;\; \dfrac{\sin 120^\circ}{q} = \dfrac{8}{10} \\[6ex] H.\;\; \dfrac{\sin 10^\circ}{q} = \dfrac{\sin 8^\circ}{120} \\[6ex] J.\;\; 120^2 = 8^2 + 10^2 - 2(8)(10) \cos q \\[4ex] K.\;\; q^2 = 8^2 + 10^2 - 2(8)(10) \cos 120^\circ \\[4ex] $
To determine the length of q, we have to use the Cosine Law beause the triangle has two sides and an included angle.
$ q^2 = r^2 + p^2 - 2(r)(p) \cos Q...\text{Cosine Law} \\[4ex] q^2 = 8^2 + 10^2 - 2(8)(10) \cos 120^\circ $
One travels at 40 miles per hour and the other at 60 miles per hour.
If they continue traveling at these constant rates, after about how many hours will they be 200 miles apart?
$ A.\;\; 1.4 \\[3ex] B.\;\; 2.8 \\[3ex] C.\;\; 3.2 \\[3ex] D.\;\; 7.7 \\[3ex] E.\;\; 8.7 \\[3ex] $
The diagram representing the information is shown:
$ e^2 = 60^2 + 40^2 ...Pythagorean\;\;Theorem \\[3ex] e^2 = 3600 + 1600 \\[3ex] e^2 = 5200 \\[3ex] e = \sqrt{5200} \\[3ex] e = 72.11102551\;mph \\[3ex] But: \\[3ex] time = \dfrac{distance}{speed} \\[5ex] distance = 200\;miles \\[3ex] speed = 72.11102551\;mph \\[3ex] time = \dfrac{200}{72.11102551} \\[5ex] time = 2.773500981 \\[3ex] time \approx 2.8\;seconds $
Point D is on $\overline{BC}$, $m\angle ADC = 60^\circ,\;\;m\angle ABC = 30^\circ$, and BD = 100 feet.
What is the length, in feet, of $\overline{AC}$
$ A.\;\; 50 \\[3ex] B.\;\; \dfrac{100}{3} \\[5ex] C.\;\; \dfrac{200}{3} \\[5ex] D.\;\; 50\sqrt{3} \\[3ex] E.\;\; 200\sqrt{3} \\[3ex] $
$ Let: \\[3ex] |AC| = y \\[3ex] |CD| = x \\[5ex] \underline{\triangle ACD} \\[3ex] \angle CAD + \angle ADC + \angle ACD = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[3ex] \angle CAD + 60 + 90 = 180 \\[3ex] \angle CAD = 180 - 60 - 90 \\[3ex] \angle CAD = 30^\circ \\[5ex] \dfrac{x}{\sin 30^\circ} = \dfrac{y}{\sin 60^\circ}...Sine\;\;Law \\[5ex] x = \dfrac{y\sin 30}{\sin 60} \\[5ex] \underline{\triangle ACB} \\[3ex] \angle CAB + \angle ABC + \angle ACB = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[3ex] \angle CAB + 30 + 90 = 180 \\[3ex] \angle CAB = 180 - 30 - 90 \\[3ex] \angle CAB = 60^\circ \\[5ex] \angle CAB = \angle CAD + \angle DAB ...diagram \\[3ex] 60 = 30 + \angle DAB \\[3ex] \angle DAB = 60 - 30 \\[3ex] \angle DAB = 30^\circ \\[5ex] \dfrac{x + 100}{\sin 60^\circ} = \dfrac{y}{\sin 30^\circ} ...Sine\;\;Law \\[5ex] x + 100 = \dfrac{y\sin 60}{\sin 30} \\[5ex] x = \dfrac{y\sin 60}{\sin 30} - 100 \\[5ex] $ This is the updated diagram
$ x = x \implies \\[3ex] \dfrac{y\sin 30}{\sin 60} = \dfrac{y\sin 60}{\sin 30} - 100 \\[5ex] y * \sin 30 * \dfrac{1}{\sin 60} = y * \sin 60 * \dfrac{1}{\sin 30} - 100 \\[5ex] \sin 30^\circ = \dfrac{1}{2}...Special\;\angle s \\[5ex] \dfrac{1}{\sin 30^\circ} = \dfrac{2}{1} = 2 \\[5ex] \sin 60^\circ = \dfrac{\sqrt{3}}{2}...Special\;\angle s \\[5ex] \dfrac{1}{\sin 60^\circ} = \dfrac{2}{\sqrt{3}} \\[5ex] \implies \\[3ex] y * \dfrac{1}{2} * \dfrac{2}{\sqrt{3}} = y * \dfrac{\sqrt{3}}{2} * \dfrac{2}{1} - 100 \\[5ex] \dfrac{y}{\sqrt{3}} = y\sqrt{3} - 100 \\[5ex] \sqrt{3}\left(\dfrac{y}{\sqrt{3}}\right) = \sqrt{3}\left(y\sqrt{3}\right) - \sqrt{3}(100) \\[5ex] y = 3y - 100\sqrt{3} \\[3ex] 100\sqrt{3} = 3y - y \\[3ex] 2y = 100\sqrt{3} \\[3ex] y = \dfrac{100\sqrt{3}}{2} \\[5ex] y = 50\sqrt{3} $
The angle of elevation from where Josiah is standing to the top of the cliff is 50°, as shown below.
Which of the following values is closest to the height, in feet, of the cliff?
(Note: $\sin 50^\circ = 0.8;\;\;\cos 50^\circ = 0.6;\;\;\tan 50^\circ = 1.2$)
$ A.\;\; 12 \\[3ex] B.\;\; 13 \\[3ex] C.\;\; 18 \\[3ex] D.\;\; 25 \\[3ex] E.\;\; 60 \\[3ex] $
$ \tan 50^\circ = \dfrac{height}{15} ...SOHCAHTOA \\[6ex] 15\tan 50 = height \\[3ex] height = 15 * \tan 50 \\[3ex] height = 15(1.2) \\[3ex] height = 18\;ft $
About how many yards less would he walk if he could walk directly from the back door to the barn door?
$ A.\;\; 8 \\[3ex] B.\;\; 19 \\[3ex] C.\;\; 23 \\[3ex] D.\;\; 26 \\[3ex] E.\;\; 31 \\[3ex] $
Let us draw the diagram to represent the information.
Walking 8 yards east means walking 8 yards due/directly east
This question is trying to assess our knowledge of the Pythagorean Theorem, as well as the importance of the hypotenuse compared to the two legs of a right triangle.
$ c^2 = 6^2 + 8^2 ...Pythagorean\;\;Theorem \\[3ex] c^2 = 36 + 64 \\[3ex] c^2 = 100 \\[3ex] c = \sqrt{100} \\[3ex] c = 10\;yards \\[5ex] Also: \\[3ex] d^2 = 5^2 + 12^2 ...Pythagorean\;\;Theorem \\[3ex] d^2 = 25 + 144 \\[3ex] d^2 = 169 \\[3ex] d = \sqrt{169} \\[3ex] d = 13\;yards \\[5ex] \underline{Distance\;\;walked\;\;by\;\;Loto} \\[3ex] = 8 + 6 + 12 + 5 \\[3ex] = 31\;yards \\[5ex] \underline{Distance\;\;that\;\;Loto\;\;could\;\;have\;\;walked\;\;directly} \\[3ex] = c + d \\[3ex] = 10 + 13 \\[3ex] = 23\;yards \\[5ex] \underline{How\;\;many\;\;yards\;\;less} \\[3ex] = Difference \\[3ex] = 31 - 23 \\[3ex] = 8\;yards $
Which of the following expressions gives QR, in meters?
(Note: For a triangle with sides of length a, b, and c that are opposite angles $\angle A$, $\angle B$, $\angle C$, respectively, $\dfrac{\sin \angle A}{a} = \dfrac{\sin \angle B}{b} = \dfrac{\sin \angle C}{c}$)
$ F.\;\; \dfrac{110\sin 40^\circ}{\sin 60^\circ} \\[5ex] G.\;\; \dfrac{110\sin 40^\circ}{\sin 80^\circ} \\[5ex] H.\;\; \dfrac{110\sin 60^\circ}{\sin 40^\circ} \\[5ex] J.\;\; \dfrac{110\sin 80^\circ}{\sin 40^\circ} \\[5ex] K.\;\; \dfrac{110\sin 80^\circ}{\sin 60^\circ} \\[5ex] $
$ \angle Q + 40^\circ + 80^\circ = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[4ex] \angle Q = 180 - 40 - 80 \\[3ex] \angle Q = 60^\circ \\[5ex] \dfrac{QR}{\sin \angle P} = \dfrac{PR}{\sin \angle Q} ...Sine\;\;Law \\[5ex] \dfrac{QR}{\sin 40^\circ} = \dfrac{110}{\sin 60^\circ} \\[5ex] QR = \dfrac{110\sin 40^\circ}{\sin 60^\circ} $
At a certain moment, the kite string forms an angle of elevation of 75° from point A on the ground.
At the same moment, the angle of elevation of the kite at point B, 240 ft from A on level ground, is 45°.
What is the length, in feet, of the string?
$ A.\;\; 60\sqrt{3} \\[3ex] B.\;\; 80\sqrt{6} \\[3ex] C.\;\; 144 \\[3ex] D.\;\; 180 \\[3ex] E.\;\; 240 \\[3ex] $
Let the length of string = string
$ \underline{\triangle AKB} \\[3ex] \angle KAB + \angle KBA + \angle AKB = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;\triangle AKB \\[3ex] 75 + 45 + \angle AKB = 180 \\[3ex] 120 + \angle AKB = 180 \\[3ex] \angle AKB = 180 - 120 \\[3ex] \angle AKB = 60^\circ \\[3ex] \dfrac{string}{\sin 45^\circ} = \dfrac{240}{\sin 60^\circ} ...Sine\;\;Law \\[5ex] \sin 45 = \dfrac{\sqrt{2}}{2} ...Special\;\;Angles \\[5ex] \sin 60 = \dfrac{\sqrt{3}}{2} ... Special\;\;Angles \\[3ex] \implies \\[3ex] string \div \sin 45 = 240 \div \sin 60 \\[3ex] string \div \dfrac{\sqrt{2}}{2} = 240 \div \dfrac{\sqrt{3}}{2} \\[5ex] string * \dfrac{2}{\sqrt{2}} = 240 * \dfrac{2}{\sqrt{3}} \\[5ex] \dfrac{\sqrt{2}}{2} * string * \dfrac{2}{\sqrt{2}} = \dfrac{\sqrt{2}}{2} * 240 * \dfrac{2}{\sqrt{3}} \\[5ex] string = \dfrac{240\sqrt{2}}{\sqrt{3}} \\[5ex] = \dfrac{240\sqrt{2}}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\[5ex] = \dfrac{240\sqrt{6}}{3} \\[5ex] = 80\sqrt{6} $
$ F.\;\; \dfrac{3}{5} \\[5ex] G.\;\; \dfrac{5}{3} \\[5ex] H.\;\; \dfrac{3}{\sqrt{34}} \\[5ex] J.\;\; \dfrac{5}{\sqrt{34}} \\[5ex] K.\;\; \dfrac{\sqrt{34}}{5} \\[5ex] $
$ \text{SOH CAH TOA} \\[3ex] \tan A = \dfrac{opp}{adj} = \dfrac{5}{3} $
On the map, the detour goes 4 miles straight north, 1 mile straight west, 2 miles straight north, 6 miles straight west, 3 miles straight south, 1 mile straight east, and finally 3 miles straight south, back to the highway.
According to the map, how many more miles will a westbound driver travel by taking the detour than he would if he could stay on the highway?
$ F.\;\; 20 \\[3ex] G.\;\; 14 \\[3ex] H.\;\; 13 \\[3ex] J.\;\; 12 \\[3ex] K.\;\; 6 \\[3ex] $
Let us represent the information on a diagram
$ Total\;\;distance\;\;using\;\;the\;\;detour \\[3ex] = 4 + 1 + 2 + 6 + 3 + 1 + 3 = 20\;miles \\[3ex] Distance\;\;covered\;\;directly:\;\;highway\;\;to\;\;highway = 1 + 6 - 1 = 6\;miles \\[3ex] Difference = 20 - 6 = 14\;miles \\[3ex] $ The westbound driver will travel 14 more miles by taking the detour than he would if he could stay on the highway.
What is the measure of $\angle A$?
$ A.\;\; 43.25^\circ \\[4ex] B.\;\; 46.75^\circ \\[4ex] C.\;\; 60^\circ \\[4ex] D.\;\; 86.5^\circ \\[4ex] E.\;\; 93.5^\circ \\[4ex] $
Let us represent the information diagrammatically
$ Let\;\;\angle A = \angle C = p^\circ ...congruent\;\;\angle s \\[4ex] p + p + 93.5 = 180^\circ ...sum\;\;of\;\;\angle s\;\;of\;\;a\;\;\triangle \\[4ex] 2p = 180 - 93.5 \\[3ex] 2p = 86.5 \\[3ex] p = \dfrac{86.5}{2} \\[5ex] p = 43.25^\circ \\[4ex] \angle A = 43.25^\circ $
What is the measure of $\angle DBE$?
$ A.\;\; 19^\circ \\[4ex] B.\;\; 29^\circ \\[4ex] C.\;\; 33^\circ \\[4ex] D.\;\; 35^\circ \\[4ex] E.\;\; 54^\circ \\[4ex] $
$ \underline{\text{Based on the Diagram}} \\[3ex] \angle ABC = \angle ABE + \angle EBC \\[3ex] 87 = 68 + \angle EBC \\[3ex] \angle EBC = 87 - 68 \\[3ex] \angle EBC = 19^\circ \\[5ex] \angle DBC = \angle DBE + \angle EBC \\[3ex] \angle DBE = \angle DBC - \angle EBC \\[3ex] \angle DBE = 52 - 19 \\[3ex] \angle DBE = 33^\circ $
