3rd one plzzz..... answer it as soon as possible
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hey mate
here's the solution
here's the solution
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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
<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
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