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The angles of elevation of the top of a tower from two points

Question

The angles of elevation of the top of a tower from two points at a distance of $4 m$ and $9 m$ from the base of the tower and in the same straight line with it complementary. prove that the height of the tower is $6 m$ .

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The angles of elevation of the top of a tower from two points at a distance of

4 m

and

9 m

from the base of the tower and in the same straight line with it complementary. prove that the height of the tower is

6 m

13 mins ago

Discuss this question LIVE

13 mins ago

Text solutionVerified

Given,

θ + ϕ = \frac{π}{2} = 9 0^{\circ}

△ A BC

tan θ = \frac{A B}{BC} = \frac{h}{4}

-------(1)

△ A B D

tan ϕ = \frac{A B}{B D} = \frac{h}{9}

tan (\frac{ϕ}{2} - θ) = \frac{h}{9}

\Rightarrow cot θ = \frac{h}{9}

------------(2)

Multiplying (1) & (2), we get

tan θ \cdot cot θ = \frac{h ^{2}}{4 \times 9} = \frac{h ^{2}}{36}

∴ \Rightarrow tan θ \cdot \frac{1}{tan θ} = \frac{h ^{2}}{36}

\Rightarrow h^{2} = 36

\Rightarrow h = 6 m e t er

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Question Text	The angles of elevation of the top of a tower from two points at a distance of $4 m$ and $9 m$ from the base of the tower and in the same straight line with it complementary. prove that the height of the tower is $6 m$ .
Answer Type	Text solution:1
Upvotes	150