Question

3rd one plzzz..... answer it as soon as possible

Answer 1

hey mate
here's the solution

Answer 2

from angle sum property in ∆AQP
<Q+<A+<P=180°
90+60+<P=180°
<APQ=30°

<BPQ=<BPA+<APQ
<P=45°

Now angle of depression
<B=180°-90°-45° using angle sum property in ∆BPS
<B=45°=ANGLE OF DEPRESSION



HOPE IT HELPS