A straight highway leads to the foot of a tower ( $A B$ ) of height $50 m$ . From the top of the tower, the angles of depression of two cars, $C a n d D$ , standing on the highway are $30^{o} a n d 60^{o}$ .
Find distance of car C from the tower.

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Solution

$I n △ A B C, sin 30^{\circ} = \frac{A B}{A C} A C = 50 \times 2 = 100 m$

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A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^{\circ}$ , which is approaching the foot of the tower with a uniform speed. 12 seconds later, the angle of depression of the car is found to be $60^{\circ}$ . Find the time in seconds taken by the car to reach the foot of the tower from this point.

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A straight highway leads to the foot of a tower (AB) of height 50 m. From the top of the tower, the angles of depression of two cars, C and D, standing on the highway are 30oand60o.Find distance of car C from the tower.

In△ABC,sin30∘=ABACAC=50×2=100m

A straight highway leads to the foot of a tower ( $A B$ ) of height $50 m$ . From the top of the tower, the angles of depression of two cars, $C a n d D$ , standing on the highway are $30^{o} a n d 60^{o}$ .
Find distance of car C from the tower.

$I n △ A B C, sin 30^{\circ} = \frac{A B}{A C} A C = 50 \times 2 = 100 m$