A person standing on the top of a building of height 60 \\sqrt{3} feet observed the top of a towerto lie at an elevation of 45^\circ. That person descended to the bottom of the building and found that the top of the same tower is now at an angle of elevation of 60^\circ. The height of the tower (in feet) is

KVPY 2020-21 SB SX Exam Previous Papers

Question : 5 of 120

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A person standing on the top of a building of height

60 \√{3}

feet observed the top of a towerto lie at an elevation of

45°

. That person descended to the bottom of the building and found that the top of the same tower is now at an angle of elevation of

60°

. The height of the tower (in feet) is

Solution:

{60 \√{3}+{d}}/{{d}}=\√{3}

\⇒{d}={60 \√{3}}/{\√{3}-1}

⇒ {h} =60 \√{3}(1+{1}/{\√{3}-1})

={60 \×3}/{\√{3}-1}=90(\√{3}+1)

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