Chapter 8.2, Problem 39AYU

To determine

What is the distance from $P$ to $Q$ .

Answer to Problem 39AYU

  $\boxed{1490.48ft}$

Explanation of Solution

Given information:

To find the length of the span of a proposed ski lift from $P$ to $Q$ , a survey measures $\angle DPQ$ to be $25°$ and then walks off a distance of $1000ft$ to $R$ and measures $\angle PRQ$ to be $15°$ .What is the distance from $P$ to $Q$ .

  

Calculation:

Consider the figure, we know that,

  $\angle DPQ+\angle RPQ=180°$

  $\angle DPQ=25°$ so calculate $\angle RPQ$

  $\begin{array}{l}25°+\angle RPQ=180°\\ \angle RPQ=180°-25°\\ \angle RPQ=155°\end{array}$

Now the sum of angles in the triangle is $180°$ in triangle $RQP$

  $\begin{array}{l}\angle RQP+\angle RPQ+\angle PRQ=180°\\ \angle QRP=15°\left(given\right)\\ \angle RQP+155°+15°=180°\\ \angle RQP=180°-155°-15°\\ \angle RQP=10°\end{array}$

Now apply law of sine,

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

Now calculate the distance from $P$ to $Q$ using equality, $\frac{\sin A}{a}=\frac{\sin B}{b}$ now

From the figure,

  $\begin{array}{l}\angle A=\angle RQP,\angle B=\angle PRQ,a=PQ,b=PR\\ \frac{\sin RQP}{PR}=\frac{\sin PRQ}{PQ}\end{array}$

Substitute $\angle RQP=10°$ and $\angle PRQ=15°$

  $\begin{array}{l}\frac{\sin 10°}{1000}=\frac{\sin 15°}{PQ}\\ \frac{0.1736}{1000}=\frac{0.2588}{PQ}\\ 0.0000174=\frac{0.2588}{PQ}\\ PQ\left(0.0000174\right)=0.2588\\ PQ=\frac{0.2588}{0.0000174}\\ PQ=1490.48\end{array}$

Hence, the distance from $P$ to $Q$ is $\boxed{1490.48ft}$