Chapter 4.3, Problem 11E

To determine

The exact values of the six trigonometric functions of the angle $\theta$ for the two triangles and explain why the values are equal.

Answer to Problem 11E

The exact values of the six trigonometric functions of the angle $\theta$ in both triangles are

  $\begin{array}{l}\sin \theta =\frac{1}{3}\\ \cos \theta =\frac{2\sqrt{2}}{3}\\ \tan \theta =\frac{1}{2\sqrt{2}}\\ \cos ec\theta =3\\ \sec \theta =\frac{3}{2\sqrt{2}}\\ \cot \theta =2\sqrt{2}\end{array}$

The trigonometric functions are same for similar triangles.

Explanation of Solution

Given info:

  

Formula used:

  ${\text{Hypotenuse}}^{\text{2}}{\text{=Base}}^{\text{2}}{\text{+Perpendicular}}^{\text{2}}$

Calculation:

Using Pythagorean theorem

For triangle 1:-

  $\begin{array}{l}{\text{Hypotenuse}}^{\text{2}}{\text{=Base}}^{\text{2}}{\text{+Perpendicular}}^{\text{2}}\\ {\text{3}}^{\text{2}}={\text{Base}}^{\text{2}}+{\text{1}}^{\text{2}}\\ \text{Base}=\sqrt{8}=2\sqrt{2}\end{array}$

We have from the given drawing

  $\begin{array}{l}\sin \theta =\frac{\text{Opposite side}}{\text{Hypotenuse}}=\frac{1}{3}\\ \cos \theta =\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{2\sqrt{2}}{3}\\ \tan \theta =\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{1}{2\sqrt{2}}\\ \cos ec\theta =\frac{\text{Hypotenuse}}{\text{Opposite side}}=\frac{3}{1}=3\\ \sec \theta =\frac{\text{Hypotenuse}}{\text{Adjacent side}}=\frac{3}{2\sqrt{2}}\\ \cot \theta =\frac{\text{Adjacent side}}{\text{Opposite side}}=\frac{2\sqrt{2}}{1}=2\sqrt{2}\end{array}$

For triangle 2:-

  $\begin{array}{l}{\text{Hypotenuse}}^{\text{2}}{\text{=Base}}^{\text{2}}{\text{+Perpendicular}}^{\text{2}}\\ {\text{6}}^{\text{2}}={\text{Base}}^{\text{2}}+{\text{2}}^{\text{2}}\\ \text{Base}=\sqrt{32}=4\sqrt{2}\end{array}$

We have from the given drawing

  $\begin{array}{l}\sin \theta =\frac{\text{Opposite side}}{\text{Hypotenuse}}=\frac{2}{6}=\frac{1}{3}\\ \cos \theta =\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{4\sqrt{2}}{6}=\frac{2\sqrt{2}}{3}\\ \tan \theta =\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{2}{4\sqrt{2}}=\frac{1}{2\sqrt{2}}\\ \cos ec\theta =\frac{\text{Hypotenuse}}{\text{Opposite side}}=\frac{6}{2}=3\\ \sec \theta =\frac{\text{Hypotenuse}}{\text{Adjacent side}}=\frac{6}{4\sqrt{2}}=\frac{3}{2\sqrt{2}}\\ \cot \theta =\frac{\text{Adjacent side}}{\text{Opposite side}}=\frac{4\sqrt{2}}{2}=2\sqrt{2}\end{array}$

Here the two triangles are similar, hence the values of trigonometric functions are same for both the triangles.

  $\frac{2}{1}=\frac{4\sqrt{2}}{2\sqrt{2}}=\frac{6}{3}=2$

Conclusion:

Thus,the exact values of the six trigonometric functions of the angle $\theta$ in both triangles are

  $\begin{array}{l}\sin \theta =\frac{1}{3}\\ \cos \theta =\frac{2\sqrt{2}}{3}\\ \tan \theta =\frac{1}{2\sqrt{2}}\\ \cos ec\theta =3\\ \sec \theta =\frac{3}{2\sqrt{2}}\\ \cot \theta =2\sqrt{2}\end{array}$

The trigonometric functions are same for similar triangles.