Question

In AQRS, the measure of ZS=90°, QR = 71 feet, and SQ = 50 feet. Find the measure of ZR to the nearest degree.

Answer 1

we get that:

$\sin x=\frac{11}{32}\rightarrow x=\arcsin \frac{11}{32}=20.10^{\circ}\approx20^{\circ}^{}$