Chapter 6, Problem 44CLT

To determine

Todetermine:The angle of elevation of star.

Answer to Problem 44CLT

  ${22.6}^0$ .

Explanation of Solution

Given information:

A star is aligned the star and the top of the backboard of the basketball hoop that is 5 feet higher than eyes of observer. Horizontal distance from the backboard is 12 feet.

Formulaused:

In triangle ABC, the trigonometric ratio $\tan \theta =\frac{b}{a}$ .

Calculations:

Let $\angle PQR=\alpha$ be the angle of elevation of the star. The view of the star is shown in Figure-1.

In triangle PQR, $\tan \alpha =\frac{PR}{QR}$

  $\Rightarrow \tan \alpha =\frac{5}{12}$

  $\Rightarrow \alpha ={\tan }^{-1}\left(\frac{5}{12}\right)$

  $\Rightarrow \alpha ={\tan }^{-1}\left(\frac{5}{12}\right)={22.6}^0$ .

Thus, the angle of elevation of the star is ${22.6}^0$ .

Conclusion:

Therefore, the angle of elevation of the star is ${22.6}^0$ .