Chapter 4.3, Problem 12E

To determine

To find: The exact values of the six trigonometric functions of the angle $\theta$ for each of the triangles.

Answer to Problem 12E

The exact values of the six trigonometric functions of the angle $\theta$ for each of the triangles are

  $\sin \theta =\frac{1.5}{2.5},\cos \theta =\frac{2}{2.5},\mathrm{Sec}\theta =\frac{2.5}{2},\csc \theta =\frac{2.5}{1.5},\tan \theta =\frac{1.5}{2},\cot \theta =\frac{2}{1.5}$

The function values of each triangles are same because the triangles are similar triangle.

Explanation of Solution

Given information:

  

Calculation: From the properties of trigonometric function,

  $\begin{array}{l}\sin \theta =\frac{\text{Perpendicular}}{\text{Hypotenuse}},\text{ csc}\theta \text{=}\frac{\text{Hypotenuse}}{\text{Perpendicular}}\\ \cos \theta =\frac{\text{Base}}{\text{Hypotenuse}},\text{ sec}\theta \text{=}\frac{\text{Hypotenuse}}{\text{Base}}\\ \tan \theta =\frac{\text{Perpendicular}}{\text{Base}},\text{ cot}\theta =\frac{\text{Base}}{\text{Perpendicular}}\end{array}$   

For the first triangle, Perpendicular = $6$ and Hypotenuse = $15$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 15=\sqrt{{\left(6\right)}^2+{\left(\text{Base}\right)}^2}\\ 225=36+{\text{Base}}^2\\ \text{Base}=\sqrt{225-36}\\ \text{Base}=\sqrt{189}\end{array}$

Hence, The third side of the triangle is, Perpendicular $=\sqrt{189}$

Let’s find the six trigonometric functions,

  $\begin{array}{l}\sin \theta =\frac{6}{15}=\frac{1}{2.5}\\ \cos \theta =\frac{\sqrt{189}}{15}\\ \mathrm{Sec}\theta =\frac{15}{\sqrt{189}}\\ \csc \theta =2.5\\ \tan \theta =\frac{6}{\sqrt{189}}\\ \cot \theta =\frac{\sqrt{189}}{6}\end{array}$

For the second triangle, Perpendicular = $4$ and Hypotenuse = $10$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 10=\sqrt{{\left(4\right)}^2+{\left(\text{Base}\right)}^2}\\ 100=16+{\text{Base}}^2\\ \text{Base}=\sqrt{100-16}\\ \text{Base}=\sqrt{84}\end{array}$

Hence, The third side of the triangle is, Perpendicular $=\sqrt{84}$

Let’s find the six trigonometric functions,

  $\begin{array}{l}\sin \theta =\frac{1}{2.5}\\ \cos \theta =\frac{\sqrt{84}}{10}\\ \mathrm{Sec}\theta =\frac{10}{\sqrt{84}}\\ \csc \theta =2.5\\ \tan \theta =\frac{4}{\sqrt{84}}\\ \cot \theta =\frac{\sqrt{84}}{4}\end{array}$

Hence , The exact values of the six trigonometric functions of the angle $\theta$ for each of the triangles are

  $\sin \theta =\frac{1}{2.5},\cos \theta =\frac{\sqrt{84}}{10},\mathrm{Sec}\theta =\frac{10}{\sqrt{84}},\csc \theta =2.5,\tan \theta =\frac{4}{\sqrt{84}},\cot \theta =\frac{\sqrt{84}}{4}$

The function values of each triangles are same because the triangles are similar triangle.