Selected Trigonometry Exercises

Exercise 51:

The two equal sides of an isoscles triangle are each 42 inches. if the base neasures 30 inches, waht is the height? what is the measure of the two equal angles?

Solution:

From the triangle we know that BD bisects AC

This yields that BD=15.

We simply have:
$h=\sqrt{42^{2}-15^{2}}=39.2$

The height is:
$h=39.2\; inches$

Finding anle in A or C:
$\cos {A}=\frac{15}{42}$

This yields:
$m\angle A=m\angle B\approx 69^{\circ}$

Exercise 52:

An equilateral triangle has an altitude of 4.33 inches. What is the length of the sides?

Solution:

From the graph we can see that:

$s^{2}=(4.33)^{2}+(\frac{s}{2})^{2}$
$s^{2}-\frac{s^{2}}{4}=(4.33)^{2}$
$\frac{3s^{2}}{4}=(4.33)^{2}$
$\frac{s\sqrt{3}}{2}=4.33$
$s=\frac{4.33 \times 2}{\sqrt{3}}\approx 5$

Finally:
The side $s=5\;inches$

Exercise 53:

From a point on the ground, the angle of inclination to the top of a radio antenna on the top of a building is $47^{\circ}30'$ . Moving 33 feet farther from the building, the angle of inclination is now $42^{\circ}10'$ . How high off the ground is the top of the radio antenna?

Solution:

From the figure:
$\tan \alpha=\frac{h}{AD}$
$\tan \beta=\frac{h}{AD+33}$

We get:
$AD=\frac{h}{\tan \alpha}$
$AD+33=\frac{h}{\tan \beta}$

If we substract:
$AD+33-AD=\frac{h}{\tan \beta}-\frac{h}{\tan \alpha}$
$h(\frac{1}{\tan \beta}-\frac{1}{\tan \alpha})=33$
$0.18780524h=33$
$h=175.7$

Finally:
$h\approx 176\;feet$

Exercise 54:

A lady standing 5 feet from a mirror notices that the angle of depression from her eyes to the bottom of the mirror is $12^{\circ}$ , while the angle of elevation to the top of the mirror is $11^{\circ}$ . Find the vertical dimension of the mirror.

Solution:

From the image:
The Dimension of the mirror:
$BC=BD+DC$
$\tan 11^{\circ}=\frac{BD}{5}$
$\tan 12^{\circ}=\frac{DC}{5}$

We get:
$BD=5\tan 11^{\circ}$
$DC=5\tan 12^{\circ}$
$BD+DC=5(\tan 11^{\circ}+\tan 12^{\circ})$
$BC=2.03\;feet$

Finally,the Dimension of the mirror:
$BC=2.03\;feet$

Exercise 55:

A tree on one side of a river is due west of a rock on the other side of the river. From a stake 21 yards north of the rock, the bearing of the tree is $198^{\circ}.2$ . How far is the tree from the rock?

Solution:

$198^{\circ}.2=S 18.2^{\circ}W$
From the figure:
$\tan 18.2^{\circ}=\frac{d}{21}\Rightarrow d=21 \tan 18.2^{\circ}\approx 7\;yards$

The distance is $\approx 7\;yards$

Exercise 56:

A ship travels on a course $322^{\circ}50'$ for 79.5 miles. How many miles north and how many miles west did it travel?

Solution:

From Figure:
$\sin {52^{\circ}50'}=\frac{y}{79.5}\Rightarrow y=79.5\sin {52^{\circ}50'}=63.4$

$\cos {52^{\circ}50'}=\frac{x}{79.5}\Rightarrow x=79.5\cos {52^{\circ}50'}=48$

Finally:
The ship travelled 63.4 miles north and 48 miles west.

Exercise 57:

A ship is steering due East in a river with a constant speed of 3 knots. The river is running North. The vessel surface speed is 12 knots. Find the ship’s true heading.

Solution:

From the graph we have:
$\tan {\beta}=\frac{3}{12}\Leftarrow \beta=14.036 ^{\circ}$

The true speed:
$AC=\sqrt{12^{2}+3^{2}}=12.37$

The true course is:
$TCo=90-\beta=75.964$

Finally:
The ship is on a True course of $076^{\circ}=N76^{\circ}E$
The ship ground speed is $12.37\;knots$

Exercise 58:

An airplane is heading $030^{\circ}$ at a speed of 200 miles per hour through the air. The air currents are moving at a constant speed of $30$ miles per hour in the direction $300^{\circ}$ . Find the True Course of this airplane.