Chapter 9.1, Problem 31PS

Solution

To find: the length of the piano prop holding open the piano.

  $63.39cm$

Given information:

The diagram representing the given scenario is as shown:

  

Definition Used:

Let $\theta$ be an acute angle of a right triangle. The six trigonometric functions of $\theta$ are defined as follows:

  $\sin \theta =\frac{opposite}{hypotenuse}$ , $\cos \theta =\frac{adjacent}{hypotenuse}$ , $\tan \theta =\frac{opposite}{adjacent}$

  $\cot \theta =\frac{adjacent}{opposite}$ , $\csc \theta =\frac{hypotenuse}{opposite}$ , $\sec \theta =\frac{hypotenuse}{adjacent}$

Explanation:

Let the length of the piano prop holding open the piano be x .

  

Now,

  $\begin{array}{c}\sin 25°=\frac{opposite}{hypotenuse}\\ \sin 25°=\frac{x}{150}\\ x=150\sin 25°\\ x\approx 63.39cm\end{array}$

So, the length of the piano prop holding open the piano is approximately 63.39 cm.