From a window 15 meter high …
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From a window 15 meter high above the ground in a street the angle of elevation and depression of the top and the foot of another house on the opposite side of the street are 30degree and 45degree resprectively show that the height of the opposite house is 23.66 (root 3=1.732
Posted by Ankush Choudhary 6 years, 1 month ago
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Sia ? 6 years, 1 month ago
Let the window be at P at a height of 15 metres above the ground and CD be the house on the opposite side of the street such that the angles of deviation of the top D of house CD as seen from P is of 30o and the angle of depression of the foot C of house CD as seen from P is of 45o.

Let h metres be the height of the house CD.
We have
QD = CD- CQ = CD -A P = ( h - 15) metres.
In △PQC, we have
tan 45o = QCPQ⇒1=15PQ⇒PQ=15 metres.
In △PQC, we have
tan30∘=QDPQ
⇒1√3=h−1515⇒h−15=15√3⇒h−15=5√3
⇒h=15+5×1.732=23.66metres.
Hence, the height of the opposite house is 23.66 metres.
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