Chapter 8.2, Problem 40AYU

To determine

Find the height $QD$ of the mountain.

Answer to Problem 40AYU

  $\boxed{1490.48ft}$

Explanation of Solution

Given information:

Find the height $QD$ of the mountain.

  

Calculation:

Consider the figure, $\Delta QDP$ is right triangle, and $\Delta PQR$ is an oblique triangle.

  $\angle PRQ=15°andPR=1000ft$ ...... $\left(1\right)$

By using property of triangle,

  $\begin{array}{l}\angle QPR+25°=180°\\ \angle QPR=180°-25°\\ \angle QPR=155°\mathrm{.....}\left(2\right)\end{array}$

Also,

  $\begin{array}{l}\angle QPR+\angle PRQ+\angle RQP=180°\\ \angle RQP=180°-\left(\angle QPR+\angle PRQ\right)\\ \angle RQP=180°-\left(155°+15°\right)from\left(1\right)\&\left(2\right)\\ \angle RQP=10°\end{array}$

Now apply law of sine,

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

  $\frac{\sin P}{QR}=\frac{\sin Q}{PR}=\frac{\sin Q}{PQ}$

Substitute the calculated values,

  $\begin{array}{l}\frac{\sin 155°}{QR}=\frac{\sin 10°}{1000}=\frac{\sin 15°}{PQ}\\ \frac{\sin 10°}{1000}=\frac{\sin 15°}{PQ}\\ PQ=1000\times \frac{\sin 15°}{10°}\\ PQ=1000\times \frac{0.25882}{0.17365}\\ PQ=1490.48498ft\end{array}$

In right $\Delta QDP$ use trigonometric ratio of acute angle for right triangle,

  $\begin{array}{l}\sin 25°=\frac{QD}{PQ}\\ \sin 25°=\frac{QD}{1490.48498}\\ QD\approx 1490.48498\times \sin 25°\\ QD\approx 1490.48498\times 0.42262\\ QD\approx 629.906ft\end{array}$

Hence, the height of the mountain is $\boxed{1490.48ft}$