the angle of elevation of top Q of a vertical tower PQ from a point X on the ground is 60° at a point Y. 40m vertically above x the angle of elevation is 45° . Find the height of the PQ.
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In ΔYRQ, we have
tan45o = YRQR
1=YRx
YR=x or XP=x [because YR=XP] ---- (1)
Now In ΔXPQ, we have
tan60o = PXPQ3 = x+40
(from equation 1)x( 3−1)=40
x=3 −140
x= 1.73−140 =54.79 m
So, the height of the tower, PQ=x+40=54.79+40=94.79 m.
Distance PX=54.79 m.
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