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QJEEM18Ma1c91: JEE Main 2018 Mathematics - Class 12 JEEM18Ma1c91 Solution: PQR is a triangular park with PQ = PR = 200 m. A T.V. tower…

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 :

Multiple Choice Options

  • A 50250\sqrt 2
  • B 100
  • C 50
  • D 1003100\sqrt 3

Step-by-Step Verified Solution

JEE Main 2018 (Offline) Mathematics - Height and Distance Question 35 English Explanation

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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