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A person observes the top of a tower from a point A on the gr

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A person observes the top of a tower from a point $A$ on the ground. The elevation of the tower from this point is $6 0^{\circ}$ . He moves $60 m$ in the direction perpendicular to the line joining $A$ and base of the tower. The angle of elevation of the tower from the point is $4 5^{\circ}$ . Then the height of the tower (in meters) is

60 \frac{3}{2}

602

603

60 \frac{2}{3}

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A person observes the top of a tower from a point

A

on the ground. The elevation of the tower from this point is

6 0^{\circ}

. He moves

60 m

in the direction perpendicular to the line joining

A

and base of the tower. The angle of elevation of the tower from the point is

4 5^{\circ}

. Then the height of the tower (in meters) is

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Text solutionVerified

Let the base of tower be

B

and the top be

T

Initial position is

A

. Let the new position be

C

BT = height of the tower = h

∠ T A B = 6 0^{\circ}

and

∠ TCB = 4 5^{\circ}

t an 6 0^{\circ} = \frac{h}{A B} \Rightarrow A B = \frac{h}{3}

t an 4 5^{\circ} = \frac{h}{CB} \Rightarrow CB = h

Applying Pythagoras theorem in

△ C A B,

C B^{2} = C A^{2} + A B^{2}

\Rightarrow h^{2} = 6 0^{2} + (\frac{h}{3})^{2} \Rightarrow h = 60 \frac{3}{2} m

So, Option (A)

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Question Text	A person observes the top of a tower from a point $A$ on the ground. The elevation of the tower from this point is $6 0^{\circ}$ . He moves $60 m$ in the direction perpendicular to the line joining $A$ and base of the tower. The angle of elevation of the tower from the point is $4 5^{\circ}$ . Then the height of the tower (in meters) is
Answer Type	Text solution:1
Upvotes	150