Question

What's the length of PS, QS, and area of PQR?

Answer 1

First, notice the following right triangle that we can get with the vertices Q,S and P:

then, we can find PS and QS using the trigonometric functions sine and cosine:

$\begin{gathered} \sin (62.58)=\frac{\text{opposite side}}{hypotenuse}=\frac{PS}{4.5} \\ \Rightarrow PS=4.5\cdot\sin (62.58)=3.99 \\ \cos (62.58)=\frac{\text{adjacent side}}{hypotenuse}=\frac{QS}{4.5} \\ \Rightarrow QS=4.5\cdot\cos (62.58)=2.07 \end{gathered}$

therefore, PS = 3.99 and QS = 2.07

Now, notice that the base of the triangle PQR is:

$QR=QS+SR$

since we already know that SR = 5, we have that the base of the triangle is:

$QR=2.07+5=7.07$

Now we can find the area of the triangle PQR:

$A_{\Delta\text{PQR}}=\frac{\text{base}\cdot\text{height}}{2}=\frac{QR\cdot PS}{2}=\frac{7.07\cdot3.99}{2}=\frac{28.21}{2}=14.11$

therefore, the area of triangle PQR is 14.11 u^2