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Height and Distance
Height and Distance
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Question

PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45∘^\circ , 30∘^\circ and 30∘^\circ , then the height of the tower (in m) is :

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Solution

Let height of tower TM=hTM = h From triangle PMT,PMT, tan⁡45o=hPM\tan {45^o} = {h \over {PM}} ⇒   PM=h \Rightarrow \,\,\,PM = h From triangle TRM,TRM, tan⁡30o=TMRM\tan {30^o} = {{TM} \over {RM}} ⇒   RM=3h \Rightarrow \,\,\,RM = \sqrt3 h From triangle. PMR,PMR, PM2+MR2=(200)2P{M^2} + M{R^2} = {\left( {200} \right)^2} ⇒   h2+(3h)2=2002 \Rightarrow \,\,\,h{}^2 + {\left( {\sqrt 3 h} \right)^2} = {200^2} ⇒   4h2=2002 \Rightarrow \,\,\,4{h^2} = {200^2} ⇒   h2=20024 \Rightarrow \,\,\,{h^2} = {{{{200}^2}} \over 4} ⇒   h=2002=100 m \Rightarrow \,\,\,h = {{200} \over 2} = 100\,m ∴   \therefore\,\,\, height of tower =100=100 m.m.

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