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Question

The angle of elevation of the top of a pillar at any point A on the ground is 15∘ . On walking 100 ft. towards the pillar, the angle becomes 30∘. Height of the pillar and its distance from A are _______ and _______ respectively.

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Solution

Let CD be the height of the pillar and AB=100 ft be the distance travelled by the man.Let initial angle of elevation ∠DAC=15∘ and final angle be ∠DBC=30∘.Consider, right ΔDAC,⇒tan15∘=DCAC=DC100+x⇒2−√3=DC100+x [∵tan15∘=2−√3]⇒(2−√3)(100+x)=DC...(i)Consider, right ΔDBC,⇒tan30∘=DCBC=DCx ...(ii)From (i), we get,⇒1√3=(2−√3)(100+x)x⇒x=√3(200+2x−100√3−√3x)⇒x=200√3+2√3x−300−3x⇒x−2√3x+3x=200√3−300⇒2√3x(2−√3)=100√3(2−√3)⇒x=50 ft.Therefore, from (ii)⇒DC=1√3×50=50√33 ft.Height of the pillar and its distance from A are 50√33 ft and 100+50=150 ft respectively.

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