Practice questions

Chapter 1: Angles and the trigonometric functions

Try each problem first. Then check your work with the solution.

1. Angles and their measure.

(a) Change ${225^\circ}$ to radians, and change ${-\dfrac{7\pi}{6}}$ to degrees.

(b) Find a positive and a negative angle coterminal with ${\dfrac{7\pi}{3}}$.

(c) A circle has radius ${8}$ in. Find the length of the arc and the area of the sector cut off by a central angle of ${\dfrac{2\pi}{3}}$.

(d) A car tire has a radius of ${14}$ inches and turns at ${600}$ revolutions per minute. How fast is the car going, in miles per hour?

Solution:

(a)

Multiply by ${\dfrac{\pi}{180}}$ to get radians:

$225\cdot\dfrac{\pi}{180}=\dfrac{225\pi}{180}=\dfrac{5\pi}{4}$

Multiply by ${\dfrac{180}{\pi}}$ to get degrees:

$-\dfrac{7\pi}{6}\cdot\dfrac{180}{\pi}=-\dfrac{7\cdot 180}{6}=-210^\circ$

(b)

A full turn is ${2\pi=\dfrac{6\pi}{3}}$. Subtract it once, then again:

$\dfrac{7\pi}{3}-\dfrac{6\pi}{3}=\dfrac{\pi}{3}$

$\dfrac{\pi}{3}-\dfrac{6\pi}{3}=-\dfrac{5\pi}{3}$

So ${\dfrac{\pi}{3}}$ and ${-\dfrac{5\pi}{3}}$ are coterminal with ${\dfrac{7\pi}{3}}$.

(c)

The angle is already in radians. Arc length:

$s=r\theta=8\cdot\dfrac{2\pi}{3}=\dfrac{16\pi}{3}\approx 16.76$ in

Sector area:

$\begin{align*}A&=\dfrac{1}{2}r^2\theta=\dfrac{1}{2}\cdot 64\cdot\dfrac{2\pi}{3}\\&=\dfrac{64\pi}{3}\approx 67.02\text{ in}^2\end{align*}$

(d)

Each revolution is ${2\pi}$ radians, so the angular speed is

${\omega=600\cdot 2\pi=1200\pi}$ radians per minute

The car moves as fast as a point on the edge of the tire:

$\begin{align*}v=r\omega&=14\cdot 1200\pi\\&=16{,}800\pi\\&\approx 52{,}779\end{align*}$

That is ${52{,}779}$ inches per minute. Multiply by ${60}$ minutes per hour, and divide by ${63{,}360}$ inches per mile:

$v\approx\dfrac{52{,}779\cdot 60}{63{,}360}\approx 50.0$

The car is going about ${50}$ miles per hour.

2. Right triangle trigonometry.

(a) A right triangle has legs ${5}$ and ${12}$. Let ${\theta}$ be the angle opposite the leg ${5}$. Find the six trigonometric functions of ${\theta}$.

(b) The angle ${\theta}$ is acute, and ${\tan\theta=2}$. Find ${\sin\theta}$ and ${\cos\theta}$.

(d) Find the acute angle ${\theta}$ with ${\sin\theta=\cos 25^\circ}$.

Solution:

(a)

Find the hypotenuse: ${\sqrt{5^2+12^2}=\sqrt{169}=13}$. From ${\theta}$, the opposite side is ${5}$ and the adjacent side is ${12}$:

${\sin\theta=\dfrac{5}{13}}$${\csc\theta=\dfrac{13}{5}}$

${\cos\theta=\dfrac{12}{13}}$${\sec\theta=\dfrac{13}{12}}$

${\tan\theta=\dfrac{5}{12}}$${\cot\theta=\dfrac{12}{5}}$

(b)

Since $\tan\theta=\dfrac{2}{1}=\dfrac{\text{opp}}{\text{adj}}$, draw a right triangle with opposite side ${2}$ and adjacent side ${1}$. The hypotenuse is ${\sqrt{2^2+1^2}=\sqrt{5}}$. So

$\sin\theta=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}$

$\cos\theta=\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}$

(c)

Use ${\sin 30^\circ=\dfrac{1}{2}}$ and ${\tan 60^\circ=\sqrt{3}}$:

$2\cdot\dfrac{1}{2}+\sqrt{3}\cdot\sqrt{3}=1+3=4$

(d)

Sine and cosine are cofunctions, so

$\begin{align*}\cos 25^\circ&=\sin(90^\circ-25^\circ)\\&=\sin 65^\circ\end{align*}$

So ${\theta=65^\circ}$.

3. Trigonometric functions of any angle.

(a) The terminal side of ${\theta}$ passes through ${(5,-12)}$. Find ${\sin\theta}$, ${\cos\theta}$, and ${\tan\theta}$.

(b) In which quadrant does ${\theta}$ lie if ${\cos\theta<0}$ and ${\cot\theta>0}$?

(d) Suppose ${\sec\theta=-3}$ and ${\theta}$ is in quadrant III. Find ${\sin\theta}$ and ${\tan\theta}$.

Solution:

(a)

Here ${x=5}$ and ${y=-12}$, so ${r=\sqrt{25+144}=13}$:

${\sin\theta=-\dfrac{12}{13}}$${\cos\theta=\dfrac{5}{13}}$

${\tan\theta=-\dfrac{12}{5}}$

(b)

Cosine is negative in quadrants II and III. Cotangent has the same sign as tangent, so it is positive in quadrants I and III. Both hold only in quadrant III.

(c)

${210^\circ}$ is in quadrant III, with reference angle ${30^\circ}$. Cosine is negative there:

$\cos 210^\circ=-\cos 30^\circ=-\dfrac{\sqrt{3}}{2}$

${135^\circ}$ is in quadrant II, with reference angle ${45^\circ}$. Tangent is negative there:

${\tan 135^\circ=-\tan 45^\circ=-1}$

${300^\circ}$ is in quadrant IV, with reference angle ${60^\circ}$. Sine is negative there, so ${\sin 300^\circ=-\dfrac{\sqrt{3}}{2}}$. Take the reciprocal:

$\csc 300^\circ=-\dfrac{2}{\sqrt{3}}=-\dfrac{2\sqrt{3}}{3}$

(d)

Cosine is the reciprocal of secant, so ${\cos\theta=-\dfrac{1}{3}}$. Use ${\sin^2\theta+\cos^2\theta=1}$:

$\begin{align*}\sin^2\theta&=1-\dfrac{1}{9}=\dfrac{8}{9}\\\sin\theta&=\pm\dfrac{\sqrt{8}}{3}=\pm\dfrac{2\sqrt{2}}{3}\end{align*}$

Sine is negative in quadrant III, so ${\sin\theta=-\dfrac{2\sqrt{2}}{3}}$. Then

$\begin{align*}\tan\theta&=\dfrac{\sin\theta}{\cos\theta}=\dfrac{-2\sqrt{2}/3}{-1/3}\\&=2\sqrt{2}\end{align*}$

4. The unit circle.

(a) Find the terminal points of ${t=\dfrac{11\pi}{6}}$ and ${t=\dfrac{3\pi}{4}}$.

(b) The point ${P\left(x,\dfrac{5}{13}\right)}$ is on the unit circle in quadrant II. Find ${\sin t}$, ${\cos t}$, and ${\tan t}$ for its number ${t}$.

(d) Is there a number ${t}$ with ${\cos t=-1.2}$? Explain.

Solution:

(a)

${\dfrac{11\pi}{6}}$ is in quadrant IV, with reference number

${2\pi-\dfrac{11\pi}{6}=\dfrac{\pi}{6}}$

In quadrant IV, ${y}$ is negative:

$P\left(\dfrac{\sqrt{3}}{2},-\dfrac{1}{2}\right)$

${\dfrac{3\pi}{4}}$ is in quadrant II, with reference number

${\pi-\dfrac{3\pi}{4}=\dfrac{\pi}{4}}$

In quadrant II, ${x}$ is negative:

$P\left(-\dfrac{\sqrt{2}}{2},\dfrac{\sqrt{2}}{2}\right)$

(b)

Use ${x^2+y^2=1}$:

${x^2=1-\dfrac{25}{169}=\dfrac{144}{169}}$, so ${x=\pm\dfrac{12}{13}}$

In quadrant II, ${x}$ is negative, so ${x=-\dfrac{12}{13}}$. Then

${\sin t=\dfrac{5}{13}}$${\cos t=-\dfrac{12}{13}}$

$\tan t=\dfrac{5/13}{-12/13}=-\dfrac{5}{12}$

(c)

Sine is odd, so

$\sin\left(-\dfrac{5\pi}{3}\right)=-\sin\dfrac{5\pi}{3}$

The number ${\dfrac{5\pi}{3}}$ is in quadrant IV, with reference number ${\dfrac{\pi}{3}}$, so

${\sin\dfrac{5\pi}{3}=-\dfrac{\sqrt{3}}{2}}$

Therefore

$\begin{align*}\sin\left(-\dfrac{5\pi}{3}\right)&=-\left(-\dfrac{\sqrt{3}}{2}\right)\\&=\dfrac{\sqrt{3}}{2}\end{align*}$

For the cosine, split off full turns: ${\dfrac{17\pi}{4}=\dfrac{\pi}{4}+4\pi}$, which is two full turns. So

$\cos\dfrac{17\pi}{4}=\cos\dfrac{\pi}{4}=\dfrac{\sqrt{2}}{2}$

(d)

No. The cosine is the ${x}$-coordinate of a point on the unit circle, so it is always between ${-1}$ and ${1}$. The number ${-1.2}$ is outside this range.

5. Solving right triangles. Round answers to two decimal places.

(a) Solve the right triangle with ${C=90^\circ}$, ${B=62^\circ}$, and ${b=15}$.

(b) A ${20}$-foot ladder leans against a wall. Its foot is ${6}$ feet from the wall. Find the angle the ladder makes with the ground, and how high up the wall it reaches.

(c) A kite is flying on ${150}$ m of straight string. The string makes an angle of ${40^\circ}$ with the ground. How high is the kite?

(d) From the top of a cliff ${60}$ m above the sea, the angles of depression of two boats are ${20^\circ}$ and ${35^\circ}$. The boats are in a straight line with the foot of the cliff. How far apart are the boats?

(e) A hiker walks ${4}$ km due east, then ${3}$ km due north. How far is she from her starting point, and what is her bearing from it?

Solution:

(a)

The other acute angle is ${A=90^\circ-62^\circ=28^\circ}$.

From ${B}$, side ${b=15}$ is opposite and side ${a}$ is adjacent. So use tangent:

$\begin{align*}\tan 62^\circ&=\dfrac{15}{a}\\a&=\dfrac{15}{\tan 62^\circ}\approx 7.98\end{align*}$

Use sine for the hypotenuse ${c}$:

$\begin{align*}\sin 62^\circ&=\dfrac{15}{c}\\c&=\dfrac{15}{\sin 62^\circ}\approx 16.99\end{align*}$

(b)

The ladder is the hypotenuse, ${20}$ ft. The distance from the wall, ${6}$ ft, is adjacent to the angle ${\theta}$ at the ground:

${\cos\theta=\dfrac{6}{20}}$, so ${\theta=\cos^{-1}(0.3)\approx 72.54^\circ}$

The height ${h}$ on the wall comes from the Pythagorean theorem:

$h=\sqrt{20^2-6^2}=\sqrt{364}\approx 19.08$ ft

(c)

The string is the hypotenuse, and the height ${h}$ is opposite the ${40^\circ}$ angle:

${h=150\sin 40^\circ\approx 96.42}$ m

(d)

From the top of the cliff, both boats are seen below the horizontal.

The angle of depression to each boat equals the angle of elevation from that boat to the top of the cliff. For each boat, the cliff is the opposite side and the boat's distance from the cliff is the adjacent side.

Far boat:

${\dfrac{60}{\tan 20^\circ}\approx 164.85}$ m

Near boat:

${\dfrac{60}{\tan 35^\circ}\approx 85.69}$ m

The distance between the boats is the difference:

${d\approx 164.85-85.69=79.16}$ m

(e)

East and north are at right angles, so the path makes a right triangle with legs ${4}$ and ${3}$:

${d=\sqrt{4^2+3^2}=5}$ km

For the bearing, measure the angle ${\theta}$ from north toward east. The east distance ${4}$ is opposite ${\theta}$, and the north distance ${3}$ is adjacent:

$\theta=\tan^{-1}\dfrac{4}{3}\approx 53.13^\circ$

She is ${5}$ km from her start, on a bearing of N ${53.13^\circ}$ E.

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