# Trigonometric Functions in a Right Triangle

In this section, we introduce the trigonometric functions using a right triangle. There are $$6$$ main trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant. For acute angles, these functions can be defined as ratios between the sides of a right triangle.

Consider a right triangle $$ABC$$ with an acute angle of $$A = \alpha.$$ The side $$b$$ between the angle $$\alpha$$ and the right angle $$C$$ is called the adjacent leg to angle $$\alpha.$$ Respectively, the other leg $$a$$ is called opposite to angle $$\alpha.$$

The sine of $$\alpha,$$ abbreviated $$\sin \alpha,$$ is the ratio of the length of the opposite leg to the length of the hypotenuse:

The cosine of $$\alpha,$$ abbreviated $$\cos \alpha,$$ is the ratio of the length of the adjacent leg to the length of the hypotenuse:

The tangent of $$\alpha,$$ abbreviated $$\tan \alpha,$$ is the ratio of the length of the opposite leg to the length of the adjacent leg:

The tangent function can be expressed in terms of the sine and cosine:

${\tan \alpha = \frac{a}{b} }={ \frac{a}{c} \cdot \frac{c}{b} }={ \frac{{\frac{a}{c}}}{{\frac{b}{c}}} }={ \frac{{\sin \alpha }}{{\cos \alpha }}.}$

If we take the reciprocals of the sine, cosine, and tangent functions, we obtain expressions for the other $$3$$ trigonometric functions.

The reciprocal of the sine function is called the cosecant. The cosecant of $$\alpha,$$ denoted $$\csc \alpha,$$ is the ratio of the length of the hypotenuse to the length of the opposite leg:

The reciprocal of the cosine function is called the secant. The secant of $$\alpha,$$ denoted $$\sec \alpha,$$ is the ratio of the length of the hypotenuse to the length of the adjacent leg:

The reciprocal of the tangent is called the cotangent. The cotangent of $$\alpha,$$ denoted $$\cot \alpha,$$ is the ratio of the length of the adjacent leg to the length of the opposite leg:

#### Example

The top of the Eiffel Tower is seen from a distance of $$d=500\,m$$ at an angle of $$\alpha = 31^\circ.$$ Find the tower’s height.

Solution.

We know the angle of elevation $$\alpha$$ and the adjacent leg $$d.$$ The tower’s height $$H$$ is the opposite leg in the right triangle. Hence,

${H = d\tan \alpha }={ 500 \times \tan {31^\circ} }={ 500 \times 0,6 }={ 300\,m.}$

## Solved Problems

Click or tap a problem to see the solution.

### 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 $$\cot \beta.$$

### Example 2

Evaluate the six trigonometric functions of the angle $$\alpha$$ 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

${\sin \alpha = \frac{a}{c} = \frac{5}{{13}},\;\;}\kern0pt{\cos \alpha = \frac{b}{c} = \frac{{12}}{{13}},}$

${\tan \alpha = \frac{a}{b} = \frac{5}{{12}},\;\;}\kern0pt{\cot \alpha = \frac{b}{a} = \frac{{12}}{{5}},}$

${\sin \beta = \frac{b}{c} = \frac{12}{{13}},\;\;}\kern0pt{\cos \beta = \frac{a}{c} = \frac{{5}}{{13}},}$

${\tan \beta = \frac{b}{a} = \frac{12}{{5}},\;\;}\kern0pt{\cot \beta = \frac{a}{b} = \frac{{5}}{{12}}.}$

### 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:

${\sin \alpha = \frac{a}{c} = \frac{9}{{15}} = \frac{3}{5},\;\;}\kern0pt{\cos \alpha = \frac{b}{c} = \frac{{12}}{{15}} = \frac{4}{5},}$

${\tan \alpha = \frac{a}{b} = \frac{9}{{12}} = \frac{3}{4},\;\;}\kern0pt{\cot \alpha = \frac{b}{a} = \frac{{12}}{{9}} = \frac{4}{3},}$

${\sec \alpha = \frac{c}{b} = \frac{15}{{12}} = \frac{5}{4},\;\;}\kern0pt{\csc \alpha = \frac{c}{a} = \frac{{15}}{{9}} = \frac{15}{3}.}$

### 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 $$\large{\frac{{{{360}^0}}}{n}}\normalsize = \large{\frac{{2\pi }}{n}}\normalsize.$$ 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

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