When a plane mirror $A B$ is placed horizontally on level ground, with $B$ at a distance of $60 m$ from the foot of a tower, the top of the tower and its image in the mirror subtends, an angle of $90^{\circ}$ at $B$ . The height of the tower is

30 m

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60 m

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90 m

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120 m

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Solution

The correct option is B $60 m$

As the top of the tower and its image in the mirror subtends, an angle of $90^{\circ}$ at $B$ .
As size of image is equal to size of object, tower exerts $45^{\circ}$ at $B$ .
So,
$tan 45^{\circ} = \frac{y}{60} \Rightarrow y = 60 m$

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When a plane mirror AB is placed horizontally on level ground, with B at a distance of 60 m from the foot of a tower, the top of the tower and its image in the mirror subtends, an angle of 90∘ at B. The height of the tower is

The correct option is B 60 m As the top of the tower and its image in the mirror subtends, an angle of 90∘ at B. As size of image is equal to size of object, tower exerts 45∘ at B. So, tan45∘=y60⇒y=60 m

When a plane mirror $A B$ is placed horizontally on level ground, with $B$ at a distance of $60 m$ from the foot of a tower, the top of the tower and its image in the mirror subtends, an angle of $90^{\circ}$ at $B$ . The height of the tower is

The correct option is B $60 m$

As the top of the tower and its image in the mirror subtends, an angle of $90^{\circ}$ at $B$ .
As size of image is equal to size of object, tower exerts $45^{\circ}$ at $B$ .
So,
$tan 45^{\circ} = \frac{y}{60} \Rightarrow y = 60 m$