Question

Find the exact values of the six trigonometric ratios of the angle$\theta$in the triangle.

Short answer

The six trigonometric ratios of an angle in the triangle are

$\begin{array}{l}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }\mathbf{=}\frac{\sqrt{\mathbf{15}}}{\mathbf{8}}\\ \mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\theta }\mathbf{=}\frac{\mathbf{7}}{\mathbf{8}}\\ \mathbf{t}\mathbf{a}\mathbf{n}\mathbf{\theta }\mathbf{=}\frac{\sqrt{\mathbf{15}}}{\mathbf{7}}\\ \mathbf{c}\mathbf{s}\mathbf{c}\mathbf{\theta }\mathbf{=}\frac{\mathbf{8}}{\sqrt{\mathbf{15}}}\\ \mathbf{s}\mathbf{e}\mathbf{c}\mathbf{\theta }\mathbf{=}\frac{\mathbf{7}}{\mathbf{8}}\\ \mathbf{c}\mathbf{o}\mathbf{t}\mathbf{\theta }\mathbf{=}\frac{\mathbf{7}}{\sqrt{\mathbf{15}}}\end{array}$

Step-by-step solution

Step 1. Given information.

Given: A right angle triangle with$\theta$ as its acute angle.
To find the six trigonometric ratios.
Identifying the sides of the right triangle.
And finding the required ratios.

Step 2. Finding adjacent side.By using Pythagoreantheorem,

$hy{p}^2=op{p}^2+ad{j}^2$

$\begin{array}{l}\text{adj}=7\\ \text{hyp}=8\\ \therefore {\text{opp}}^2=8^2-7^2\\ \therefore {\text{opp}}^2=15\\ \therefore \text{opp}=\sqrt{15}\end{array}$

Step 3. Finding the trigonometric function values.

The sine of an angle is defined as the ratio of the side opposite(perpendicular side) to that angle to the hypotenuse.

$\sin \theta =\frac{\sqrt{15}}{8}$

The cosine of an angle is defined as the ratio of the side adjacent to that angle to the hypotenuse.

$\cos \theta =\frac{7}{8}$

The tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle.

$\tan \theta =\frac{\sqrt{15}}{7}$

The cosecant angle is the reciprocal of sine

$\csc \theta =\frac{8}{\sqrt{15}}$

The secant angle is the reciprocal of cosine

$\sec \theta =\frac{7}{8}$

The cot angle is the reciprocal of tangent

$\cot \theta =\frac{7}{\sqrt{15}}$