# Precalculus

## Trigonometry # Trigonometric Functions in a Right Triangle

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Given the right triangle below, find the values of sin α, cos α, tan α, cot α, sin β, cos β, tan β, and cot β.

### Example 2

Evaluate the six trigonometric functions of the angle α shown in the figure.

### Example 3

What is the value of $$\tan 45^\circ?$$

### Example 4

What is the value of $$\cot 30^\circ?$$

### Example 5

Calculate the value of $$\sec 45^\circ.$$

### Example 6

Calculate the value of $$\csc 30^\circ.$$

### Example 7

To determine the height of the Empire State Building, two measurements are taken. At the initial point, the angle of elevation to the top of the building is $$\alpha = 45^\circ.$$ After moving closer by $$L = 187\,m,$$ the angle of elevation is $$\beta = 60^\circ.$$ Find the height $$H$$ of the building.

### Example 8

Find the area of a regular $$n\text{-sided}$$ polygon if the length of one side is $$a.$$

### Example 1.

Given the right triangle below, find the values of $$\sin \alpha,$$ $$\cos \alpha,$$ $$\tan \alpha,$$ $$\cot \alpha,$$ $$\sin \beta,$$ $$\cos \beta,$$ $$\tan \beta,$$ and $$\cot \beta.$$

Solution.

First we find the length of the hypotenuse using the Pythagorean theorem. Let $$a = 5,$$ $$b = 12.$$ Then

${a^2} + {b^2} = {c^2},\;\; \Rightarrow c = \sqrt {{a^2} + {b^2}} = \sqrt {{5^2} + {{12}^2}} = \sqrt {25 + 144} = \sqrt {169} = 13.$

Using the trigonometric ratios in a right triangle, we have

$\begin{array}{*{20}{l}} {\sin \alpha = \frac{a}{c} = \frac{5}{{13}},} &{\cos \alpha = \frac{b}{c} = \frac{{12}}{{13}},}\\[1em] {\tan \alpha = \frac{a}{b} = \frac{5}{{12}},} &{\cot \alpha = \frac{b}{a} = \frac{{12}}{{5}},}\\[1em] {\sin \beta = \frac{b}{c} = \frac{12}{{13}},} &{\cos \beta = \frac{a}{c} = \frac{{5}}{{13}},}\\[1em] {\tan \beta = \frac{b}{a} = \frac{12}{{5}},} &{\cot \beta = \frac{a}{b} = \frac{{5}}{{12}}.} \end{array}$

### Example 2.

Evaluate the six trigonometric functions of the angle $$\alpha$$ shown in the figure.

Solution.

Using the Pythagorean theorem, we find the second leg of the right triangle. Let $$b = 12,$$ $$c = 15.$$ Then

${a^2} + {b^2} = {c^2},\;\; \Rightarrow a = \sqrt {{c^2} - {b^2}} = \sqrt {{{15}^2} - {{12}^2}} = \sqrt {225 - 144} = \sqrt {81} = 9.$

Calculate the trigonometric functions:

$\begin{array}{*{20}{l}} {\sin \alpha = \frac{a}{c} = \frac{9}{{15}} = \frac{3}{5},} &{\cos \alpha = \frac{b}{c} = \frac{{12}}{{15}} = \frac{4}{5},}\\[1em] {\tan \alpha = \frac{a}{b} = \frac{9}{{12}} = \frac{3}{4},} &{\cot \alpha = \frac{b}{a} = \frac{{12}}{{9}} = \frac{4}{3},}\\[1em] {\sec \alpha = \frac{c}{b} = \frac{15}{{12}} = \frac{5}{4},} &{\csc \alpha = \frac{c}{a} = \frac{{15}}{{9}} = \frac{15}{3}.} \end{array}$

### Example 3.

What is the value of $$\tan 45^\circ?$$

Solution.

Consider an isosceles right triangle.

It has equal legs $$a = b,$$ and each of the acute angles of the triangle is equal to $$45^\circ.$$ Therefore

$\require{cancel}\tan {45^\circ} = \frac{a}{b} = \frac{\cancel{a}}{\cancel{a}} = 1.$

### Example 4.

What is the value of $$\cot 30^\circ?$$

Solution.

Consider a $$30\text{-}60\text{-}90$$ triangle, which is a special right triangle whose angles are $$30^\circ,$$ $$60^\circ,$$ and $$90^\circ.$$

In any $$30\text{-}60\text{-}90$$ triangle, the hypotenuse is twice as long as the shortest leg, so $$c = 2a.$$

Using this property together with the Pythagorean theorem, we express the other leg $$b$$ in terms of $$a:$$

${a^2} + {b^2} = {c^2},\;\; \Rightarrow b = \sqrt {{c^2} - {a^2}} = \sqrt {{{\left( {2a} \right)}^2} - {a^2}} = \sqrt {4{a^2} - {a^2}} = \sqrt {3{a^2}} = a\sqrt 3 .$

Now we can calculate the value of $$\cot 30^\circ :$$

$\cot {30^0} = \frac{b}{a} = \frac{{\cancel{a}\sqrt 3 }}{\cancel{a}} = \sqrt 3 .$

### Example 5.

Calculate the value of $$\sec 45^\circ.$$

Solution.

In an isosceles right triangle, the legs are equal to each other, that is $$a = b.$$ Then by the Pythagorean theorem,

$c = \sqrt {{a^2} + {b^2}} = \sqrt {{a^2} + {a^2}} = \sqrt {2{a^2}} = a\sqrt 2 .$

The secant of an angle of $$45^\circ$$ is given by

$\sec {45^\circ} = \sec \alpha = \frac{c}{b} = \frac{{\cancel{a}\sqrt 2 }}{\cancel{a}} = \sqrt 2 .$

### Example 6.

Calculate the value of $$\csc 30^\circ.$$

Solution.

In a $$30\text{-}60\text{-}90$$ triangle, the length of the hypotenuse is twice the length of the shortest leg. Hence $$c = 2a,$$ where the side $$a$$ is opposite to the angle $$\alpha = 30^\circ.$$

Find $$\csc 30^\circ:$$

$\csc {30^\circ} = \frac{c}{a} = \frac{{2\cancel{a}}}{\cancel{a}} = 2.$

### Example 7.

To determine the height of the Empire State Building, two measurements are taken. At the initial point, the angle of elevation to the top of the building is $$\alpha = 45^\circ.$$ After moving closer by $$L = 187\,m,$$ the angle of elevation is $$\beta = 60^\circ.$$ Find the height $$H$$ of the building.

Solution.

Let $$x$$ be the distance from the terminal point to the building. Then the initial distance is equal to $$L + x.$$ Using the trigonometric ratios for the right triangles shown in the figure, we get two equations:

$\left\{ \begin{array}{l} H = \left( {L + x} \right)\tan \alpha \\ H = x\tan \beta \end{array} \right..$

Solve this system to express $$H$$ in terms of $$L, \alpha, \text{and } \beta.$$

$\left\{ \begin{array}{l} H = \left( {L + x} \right)\tan \alpha \\ H = x\tan \beta \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} H = L\tan \alpha + x\tan \alpha \\ H = x\tan \beta \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} x\tan \beta = L\tan \alpha + x\tan \alpha \\ H = x\tan \beta \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} x\left( {\tan \beta - \tan \alpha } \right) = L\tan \alpha \\ H = x\tan \beta \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} x = \frac{{L\tan \alpha }}{{\tan \beta - \tan \alpha }}\\ H = x\tan \beta \end{array} \right.,\;\; \Rightarrow H = \frac{{L\tan \alpha \tan \beta }}{{\tan \beta - \tan \alpha }}.$

Substitute the known values and calculate the height $$H:$$

$H = \frac{{L\tan \alpha \tan \beta }}{{\tan \beta - \tan \alpha }} = \frac{{187 \cdot \tan {{45}^\circ}\tan {{60}^\circ}}}{{\tan {{60}^\circ} - \tan {{45}^\circ}}} = \frac{{187 \times 1 \times \sqrt 3 }}{{\sqrt 3 - 1}} = 187 \times 2.366 = 442\,m.$

### Example 8.

Find the area of a regular $$n\text{-sided}$$ polygon if the length of one side is $$a.$$

Solution.

A $$n\text{-sided}$$ polygon can be split up into $$n$$ triangles.

Each triangle has the central angle $$\frac{{{{360}^0}}}{n} = \frac{{2\pi }}{n}.$$ We denote half of the central angle by $$\alpha,$$ so

$\alpha = \frac{\pi }{n}.$

Find the apothem $$h:$$

$h = \frac{a}{2}\cot \alpha = \frac{a}{2}\cot \frac{\pi }{n}.$

The area of one triangle is

${A_0} = \frac{1}{2}ah = \frac{1}{2} \times a \times \frac{a}{2}\cot \frac{\pi }{n} = \frac{{{a^2}}}{4}\cot \frac{\pi }{n}.$

Then the total area of an $$n\text{-sided}$$ regular polygon is given by

$A = n{A_0} = \frac{{n{a^2}}}{4}\cot \frac{\pi }{n}.$