Question

Problem 8) In the diagram, RQ ⊥ PQ, m∠QPS = 32°, m∠RPS = 24°, and PQ = 14.Find RS to the nearest tenth of a unit. Image is provided down below. :)

Answer 1

Since S is a point on the segment QR, then:

$QS+SR=QR$

Use the diagram and trigonometric relationships to find the value of QS and QR. Then, use the above equation to solve for SR.

In the right triangle PQS, the side QS is opposite to the angle of 32º, and the side PQ is adjacent to the angle of 32º. Then, use the tangent trigonometric function to relate PQ to QS:

$\tan\left(32º\right)=\frac{QS}{PQ}$

Isolate QS and replace PQ=14:

$\begin{gathered} \Rightarrow QS=PQ\cdot\tan\mleft(32º\mright) \\ \operatorname{\Rightarrow}QS=14\tan\mleft(32º\mright) \end{gathered}$

On the other hand, notice that the side QR in the right triangle PQR is opposite to the angle QPR, which has a measure of 32º+24º=56º.

Then, by a similar reasoning, we can use again the tangent function to find QR:

$\begin{gathered} \tan\mleft(56º\mright)=\frac{QR}{PQ} \\ \Rightarrow QR=PQ\cdot\tan\mleft(56º\mright) \\ \Rightarrow QR=14\cdot\tan\mleft(56º\mright) \end{gathered}$

We have found expressions for QS and QR in terms of trigonometric relationships. Then, we can find an expression for SR:

$\begin{gathered} QS+SR=QR \\ \Rightarrow SR=QR-QS \end{gathered}$

Replace the expressions for QR and QS:

$SR=14\tan\mleft(56º\mright)-14\tan\mleft(32º\mright)$

Use a calculator to find decimal expressions for tan(56º) and tan(32º). Then, find SR:

$\begin{gathered} SR=14\times1.482560969...-14\times0.6248693519... \\ \Rightarrow SR=12.00768263... \end{gathered}$

Therefore, to the nearest tenth, the length of the side SR (or RS, which is the same) is 12.0.