The angle of depression of a car parked on the road side from the top of a 150 m high tower 30 - Find the distance of the car from the tower-

REF.ImageAB towerd distance of car from tower tan 30 =opposite sideadjacent side=ABBC 1√(3)=150d ∴ d=150√(3)m ...

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Standard IX

Mathematics

Calculating Heights and Distances

The angle of ...

Question

The angle of depression of a car parked on the road side from the top of a $150$ m high tower $30^{\circ}$ . Find the distance of the car from the tower.

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Solution

REF.Image
AB - tower
d- distance of car from tower

$tan 30 = \frac{o p p o s i t e s i d e}{a d j a c e n t s i d e} = \frac{A B}{B C}$

$\frac{1}{\sqrt{3}} = \frac{150}{d}$

$∴ d = 150 \sqrt{3} m$

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Q. Choose the correct answer of the following question:

The angle of depression of a car parked on the road from the top of a 150-m-high tower is 30°. The distance of the car from the tower is

(a)

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From the top of tower $100 m$ height, a man observes two cars on the opposite side of tower with angles of depression $30^{\circ}$ and $45^{\circ}$ respectively. Find the distance between cars.

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The angle of depression of a car parked on the road side from the top of a 150 m high tower 30∘. Find the distance of the car from the tower.

REF.ImageAB - towerd- distance of car from towertan30=opposite sideadjacent side=ABBC1√3=150d∴d=150√3m

The angle of depression of a car parked on the road side from the top of a $150$ m high tower $30^{\circ}$ . Find the distance of the car from the tower.

REF.Image
AB - tower
d- distance of car from tower

$tan 30 = \frac{o p p o s i t e s i d e}{a d j a c e n t s i d e} = \frac{A B}{B C}$

$\frac{1}{\sqrt{3}} = \frac{150}{d}$

$∴ d = 150 \sqrt{3} m$