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Height and Distance
Height and Distance
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Question

A tower PQ stands on a horizontal ground with base QQ on the ground. The point RR divides the tower in two parts such that QR=15 mQ R=15 \mathrm{~m}. If from a point AA on the ground the angle of elevation of RR is 60∘60^{\circ} and the part PRP R of the tower subtends an angle of 15∘15^{\circ} at AA, then the height of the tower is :

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Solution

For Δ\DeltaAQR, tan⁡60∘=15x\tan 60^\circ = {{15} \over x} ....... (1) From Δ\DeltaAQP, tan⁡75∘=hx\tan 75^\circ = {h \over x} ⇒(2+3)=hx \Rightarrow \left( {2 + \sqrt 3 } \right) = {h \over x} [∵\because tan⁡75∘=2+3\tan 75^\circ = 2 + \sqrt 3 ] ⇒h=(2+3)x \Rightarrow h = \left( {2 + \sqrt 3 } \right)x =(2+3)153 = \left( {2 + \sqrt 3 } \right){{15} \over {\sqrt 3 }} [From (1)] =(2+3)×1533 = \left( {2 + \sqrt 3 } \right) \times {{15\sqrt 3 } \over 3} =(2+3)×53= \left( {2 + \sqrt 3 } \right) \times 5\sqrt 3 =5(23+3) = 5\left( {2\sqrt 3 + 3} \right) m

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