Chapter 4.3, Problem 11E

To determine

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

Answer to Problem 11E

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, Base = $8$ and Hypotenuse = $10$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 10=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(8\right)}^2}\\ 100={\text{Perpendicular}}^2+64\\ \text{Perpendicular}=\sqrt{100-64}\\ \text{Perpendicular}=\sqrt{36}\\ \text{Perpendicular}=6\end{array}$

Hence, The third side of the triangle is, Perpendicular $=6$

Let’s find the six trigonometric functions,

  $\begin{array}{l}\sin \theta =\frac{6}{10}=\frac{3}{5}=\frac{1.5}{2.5}\\ \cos \theta =\frac{8}{10}=\frac{2}{2.5}\\ \mathrm{Sec}\theta =\frac{2.5}{2}\\ \csc \theta =\frac{10}{6}=\frac{5}{3}=\frac{2.5}{1.5}\\ \tan \theta =\frac{6}{8}=\frac{3}{4}=\frac{1.5}{2}\\ \cot \theta =\frac{4}{3}=\frac{2}{1.5}\end{array}$

For the second triangle, Base = $2$ and Hypotenuse = $2.5$

As per Pythagoras Theorem,

  $\begin{array}{c}\text{ Hypotenuse}=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(\text{Base}\right)}^2}\\ 2.5=\sqrt{{\left(\text{Perpendicular}\right)}^2+{\left(2\right)}^2}\\ 6.25={\text{Perpendicular}}^2+4\\ \text{Perpendicular}=\sqrt{6.25-4}\\ \text{Perpendicular}=\sqrt{2.25}\\ \text{Perpendicular}=1.5\end{array}$

Hence, The third side of the triangle is, Perpendicular $=1.5$

Let’s find the six trigonometric functions,

  $\begin{array}{l}\sin \theta =\frac{1.5}{2.5}\\ \cos \theta =\frac{2}{2.5}\\ \mathrm{Sec}\theta =\frac{2.5}{2}\\ \csc \theta =\frac{2.5}{1.5}\\ \tan \theta =\frac{1.5}{2}\\ \cot \theta =\frac{2}{1.5}\end{array}$

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

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

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