Selected Trigonometry Exercises

Selected Trigonometry Exercises

These are the selected trigonometry exercises and problems to solidify the concepts shown in this site.

Exercise 1:

A surveyor observes that at a point $A$ , located on level ground a distance 25.0 feet from the base $B$ of a flagpole, the angle between the ground and the top of the pole is $30^{\circ}$ . What is the height of the pole to the nearest hundredth of the feet.

Solution:

From the graph we don’t know the hypothenuse and we don’t need to calculate it.

We can use the tangent in this situation:

$\tan 30^{\circ}=\frac{h}{25}$

This yields:

$h=25\tan 30^{\circ}$

$h=24\times 0.57735027$

$h=14.43\; feet$

Exercise 2:

Show that the following identity is true:

$(\sec \theta+ tan \theta)(1-\sin \theta)=\cos \theta$

Solution:

From the left we can write:

(1) $\begin{equation*} \begin{split} (\sec \theta+ tan \theta)(1-\sin \theta)&=(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta})(1-\sin \theta)\\&= (\frac{1+\sin \theta}{\cos \theta})(1-\sin \theta)\\&= (\frac{1-\sin^{2} \theta}{\cos \theta})\\&=\frac{\cos^{2}\theta}{\cos \theta}\\&=\cos \theta \end{split} \end{equation*}$

Finally we can say that:

$(\sec \theta+ tan \theta)(1-\sin \theta)=\cos \theta$

Exercise 3:

The peak of Mt. Fuji in Japan is approximately12,400 feet high.

A student, several miles away, finds that the angle between the level ground and the peak is $30^{\circ}$ . How far is the student from the point on the level ground directly beneath the peak?

Solution:

$\tan 30^{\circ}=\frac{12,400}{x}$

This gives:

$x=\frac{12,400}{\tan 30^{\circ}}$

$x=\frac{12,400}{0.57735027}$

$x=21,477\;feet$

Exercise 4:

Stonehenge in Salisbury Plains, England, was constructed using solid stone blocks weighing over 99,000 pounds each. Lifting a single stone alone required 550 people, who pulled the stone up a ramp inclined at an angle of $9^{\circ}$ . What is the approximate distance a stone was moved in order to raise it to a height of 30 feet?

Solution:

Here we can see that:

$\sin 9^{\circ}=\frac{30}{x}$

This gives:

$x=\frac{30}{\sin 9^{\circ}}$

$x=\frac{30}{0.15643447}$

$x=192 \;feet$

Exercise 5:

From the top of a building that overlooks an ocean, an observer watches a boat sailing directly toward the building. If the observer is 100 feet above sea level and if the angle of depression of the boat changes from $20^{\circ}$ to $45^{\circ}$ during the period of the observation, approximate the distance travelled by the boat.

Solution:

$\tan 20^{\circ}=\frac{100}{d_{1}}$

This gives:
$d_{1}=\frac{100}{\tan 20^{\circ}}$

$\tan 45^{\circ}=\frac{100}{d_{2}}$
This yields:

$d_{2}=\frac{100}{\tan 45^{\circ}}$

The distance:
$d=d_{1}-d_{2}$
$d=\frac{100}{\tan 20^{\circ}}-\frac{100}{\tan 45^{\circ}}$
$d=175\;feet$

Exercise 6:

The highest advertising sign in the world is a large letter $I$ situated at the top of a 73-story First Interstate World Center building in Los Angeles. At a distance of 200 feet from a point directly below the sign, the angle between the ground and the to top of the sign is $78.87^{\circ}$ . What is the heght of the top of the sign?

Solution:

$\tan 78.87^{\circ}=\frac{x}{200}$

$x=200\tan 78.87^{\circ}$
$x=1,017 feet$

Exercise 7:

Simplify: $tan^{2}4\beta-sec^{2}4\beta$

Solution:

We have:

(2) $\begin{equation*} \begin{split} \tan^{2}4\beta-\sec^{2}4\beta&= (\tan 4\beta+ \sec 4\beta)(\tan 4\beta- \sec 4\beta)\\&=(\frac{\sin 4\beta}{\cos 4\beta}+\frac{1}{\cos 4\beta})(\frac{\sin 4\beta}{\cos 4\beta}-\frac{1}{\cos 4\beta})\\&=\frac{(\sin 4\beta+1)(\sin 4\beta-1)}{\cos 4\beta \cdot \cos 4\beta}\\&=\frac{\sin^{2}4\beta-1}{\cos^{2} 4\beta}\\&=\frac{-\cos^{2} 4\beta}{\cos^{2} 4\beta}\\&=-1 \end{split}\end{equation*}$

Finally:
$tan^{2}4\beta-sec^{2}4\beta=-1$

Exercise 8:

Simplify: $4tan^{2}\beta-4sec^{2}\beta$

Solution:

We have:

(3) $\begin{equation*} \begin{split} 4tan^{2}\beta-4sec^{2}\beta&=4(tan^{2}\beta-sec^{2}\beta)\\&=4(\frac{\sin^{2}\beta}{\cos^{2}\beta}-\frac{1}{\cos^{2}\beta})\\&=4(\frac{\sin^{2}\beta-1}{\cos^{2}\beta})\\&=4(\frac{-\cos^{2}\beta}{\cos^{2}\beta}\\&=-4Finally:$4\tan^{2}\beta-4\sec^{2}\beta=-4$ \end{split}\end{equation*}$

Exercise 9:

Two stars that are close may appear to be one. The ability of a telescope to separate their images is called its resolution. A smaller resolution means a better ability for the telescope to separate images in the sky. In a refracting telescope, resolution $\theta$ can be improved by using a lens with a larger diameter $D$ . The relationship between $\theta$ in degrees and $D$ in meters is gib=ven by: $\sin \theta=\frac{1.22\lambda}{D}$ , where $\lambda$ is the wavelength of light in meters. The largest refracting telescope in the world is at the University of chicago. At a wavelength of $\lambda=550 \times 10^{-9}$ meter, its resolution is $0.00003769^{\circ}$ . Find the diameter $D$ .

Solution:

From the prompt:

$\sin \theta=\frac{1.22\lambda}{D}$

We can re-write:
$D=\frac{1.22\lambda}{\sin \theta}$

We plug in the values:

$D=\frac{1.22\times 550 \times 10^{-9}}{\sin 0.00003769^{\circ}}$

Finally:
$D=1.02\; meters$

Exercise 10:

Simplify $\displaystyle{\frac{\sin^{3}\theta+ \cos^{3}\theta}{\sin \theta+ \cos \theta}}$

Solution:

(4) $\begin{equation*}\begin{split} \frac{\sin^{3}\theta+ \cos^{3}\theta}{\sin \theta+ \cos \theta}&=\frac{\sin \theta(1-\cos^{2}\theta)+\cos \theta(1-\sin^{2}\theta)}{\sin \theta+ \cos \theta}\\&=\frac{\sin \theta-\sin \theta \cos \theta \cos \theta+ \cos \theta -\cos \theta \sin \theta \sin \theta}{\sin \theta+ \cos \theta}\\&=\frac{(\sin \theta+ \cos \theta)-\sin \theta \cos \theta (\sin \theta+ \cos \theta)}{\sin \theta+ \cos \theta}\\&=1-\sin \theta \cos \theta \end{split}\end{equation*}$

Finally:

$\displaystyle{\frac{\sin^{3}\theta+ \cos^{3}\theta}{\sin \theta+ \cos \theta}}=1-\sin \theta \cos \theta$

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Selected Trigonometry Exercises