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The angles of elevation of a tower from a point on the ground

Question

The angles of elevation of a tower from a point on the ground is $3 0^{\circ}$ . At a point on the horizontal line passing through the foot of the tower and $100$ meters nearer to it. If the angle of elevation is found to be $6 0^{\circ}$ , then height of the tower is

503

meters

\frac{50}{3}

meters

1003

\frac{100}{3}

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Noah Discussed

The angles of elevation of a tower from a point on the ground is

3 0^{\circ}

. At a point on the horizontal line passing through the foot of the tower and

100

meters nearer to it. If the angle of elevation is found to be

6 0^{\circ}

, then height of the tower is

14 mins ago

Discuss this question LIVE

14 mins ago

Text solutionVerified

Let

A B = h

be the height of the tower, and let

A D = x

From right angled

△ A B D

tan 6 0^{\circ} = \frac{A B}{A D}

3 = \frac{h}{x}

x = \frac{h}{3}

From right angled

△ BC A

tan 3 0^{\circ} = \frac{A B}{C A}

\frac{1}{3} = \frac{h}{100 + x}

h = \frac{100 + x}{3} = \frac{100 + \frac{h}{3}}{3}

\frac{2}{3} h = 100

∴ h = 503 m

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Question Text	The angles of elevation of a tower from a point on the ground is $3 0^{\circ}$ . At a point on the horizontal line passing through the foot of the tower and $100$ meters nearer to it. If the angle of elevation is found to be $6 0^{\circ}$ , then height of the tower is
Answer Type	Text solution:1
Upvotes	150