Chapter 6.2, Problem 37E

To determine

To calculate: The sides of the right angled triangle.

  

Answer to Problem 37E

The triangle along with sides and acute angles is,

  

Explanation of Solution

Given information:

The right angle triangle with length of its sides and angle $\theta$ .

  

Formula used:

The trigonometric ratios for a right angle triangle are defined as,

  $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}},cos\theta =\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta =\frac{\text{opposite}}{\text{adjacent}},\csc \theta =\frac{\text{hypotenuse}}{\text{opposite}},\sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}$ and $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$ .

Calculation:

Consider the right angle triangle with length of its sides and angle $\theta$ .

  

Observe that the length of adjacent side is $106$ units.

Label the length of hypotenuse of the triangle as x and adjacent side as y .

  

Recall that the trigonometric ratios for a right angle triangle are defined as,

  $\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}},cos\theta =\frac{\text{adjacent}}{\text{hypotenuse}},\tan \theta =\frac{\text{opposite}}{\text{adjacent}},\csc \theta =\frac{\text{hypotenuse}}{\text{opposite}},\sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}$ and $\cot \theta =\frac{\text{adjacent}}{\text{opposite}}$ .

Apply it, to estimate the value of trigonometric ratios,

The value of tangent function is,

  $\begin{array}{c}\tan \frac{\pi }{5}=\frac{\text{opposite}}{\text{adjacent}}\\ \tan \frac{\pi }{5}=\frac{106}{y}\end{array}$

Divide both sides by $y$ ,

  $\begin{array}{c}y\cdot \tan \frac{\pi }{5}=\frac{106}{y}\cdot y\\ y\cdot \tan \frac{\pi }{5}=106\end{array}$

Simplify it with help of calculator,

  $\begin{array}{c}y\cdot \tan \frac{\pi }{5}=106\\ y=\frac{106}{\tan \frac{\pi }{5}}\\ y=145.9\end{array}$

Therefore, the length of adjacent side is $145.9$ units.

The value of sine function is,

  $\begin{array}{c}\sin \frac{\pi }{5}=\frac{\text{opposite}}{\text{hypotenuse}}\\ \sin \frac{\pi }{5}=\frac{106}{x}\end{array}$

Multiply both sides by $x$ ,

  $\begin{array}{c}\sin \frac{\pi }{5}=\frac{106}{x}\\ x\cdot \sin \frac{\pi }{5}=\frac{106}{x}\cdot x\end{array}$

Simplify it with help of calculator,

  $\begin{array}{c}x\cdot \sin \frac{\pi }{5}=106\\ x=\frac{106}{\sin \frac{\pi }{5}}\\ x=180.3\end{array}$

Therefore, the length of hypotenuse is $180.3$ units.

Observe that the angle A and C are complementary angles so,

  $\begin{array}{c}\angle A+\angle C=90°\\ \angle A+\frac{\pi }{5}=90°\\ \angle A=\frac{3\pi }{10}\end{array}$

Thus, the triangle so formed is provided below,