The angle of elevation of a tower from a distance of 100 mt. from its foot is 30∘. Then the height of the tower is

tanθ=BCABtan30∘=100h31=100h⇒h=3100.

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The angle of elevation of a tower from a distance of 100 mt. f

Question

The angle of elevation of a tower from a distance of $100 m t$ . from its foot is $3 0^{\circ}$ . Then the height of the tower is

\frac{50}{3} m

503 m

\frac{100}{3} m

1003 m

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The angle of elevation of a tower from a distance of

100 m t

. from its foot is

3 0^{\circ}

. Then the height of the tower is

6 mins ago

Discuss this question LIVE

6 mins ago

tan θ = \frac{A B}{BC}

tan 3 0^{\circ} = \frac{h}{100}

\frac{1}{3} = \frac{h}{100} \Rightarrow h = \frac{100}{3}

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Question Text	The angle of elevation of a tower from a distance of $100 m t$ . from its foot is $3 0^{\circ}$ . Then the height of the tower is
Answer Type	Text solution:1
Upvotes	150