The distance between two pillars is 120 text{ | vedclass

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Question

(D) Let the height of the shorter pillar $CD = h$ and the height of the taller pillar $AB = 3h$.The distance between the pillars $BD = 120 	ext{ m}$.

Vedclass · Accepted Answer

$103.92$

Vedclass · Answer

(D) Let the height of the shorter pillar $CD = h$ and the height of the taller pillar $AB = 3h$.The distance between the pillars $BD = 120 	ext{ m}$.Let $E$ be the midpoint of $BD$,so $BE = ED = 60 	ext{ m}$.In $	riangle ABE$,$	an 	heta = \frac{AB}{BE} = \frac{3h}{60} = \frac{h}{20} \quad ... (i)$In $	riangle CDE$,$	an(90^\circ - 	heta) = \frac{CD}{ED} = \frac{h}{60} \implies \cot 	heta = \frac{h}{60} \quad ... (ii)$Multiplying $(i)$ and $(ii)$:$	an 	heta \cdot \cot 	heta = \frac{h}{20} \cdot \frac{h}{60}$$1 = \frac{h^2}{1200}$$h^2 = 1200$$h = \sqrt{1200} = 20\sqrt{3} 	ext{ m}$.The height of the taller pillar is $3h = 3 	imes 20\sqrt{3} = 60\sqrt{3} 	ext{ m}$.Using $\sqrt{3} = 1.732$,height $= 60 	imes 1.732 = 103.92 	ext{ m}$.

The distance between two pillars is $120 \text{ m}$. The height of one pillar is thrice the other. The angles of elevation of their tops from the midpoint of the line connecting their feet are complementary to each other. The height (in $\text{m}$) of the taller pillar is (Use: $\sqrt{3} = 1.732$)

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