A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, angles of depressions of two cars standing on the highway are $30^{\circ}$ and $60^{\circ}$ . what is the distance between the cars. ?

43.3 m

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57.73 m

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86.6 m

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100 m

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Solution

The correct option is A $57.73 m$
Height of tower = $50$ m
Let the distance between car 1 and the tower = $d_{1}$
Let the distance between car 2 and the tower = $d_{2}$
Thus, $tan 30 = \frac{50}{d_{1}}$
$d_{1} = 50 \sqrt{3}$
$tan 60 = \frac{50}{d_{2}}$
$d_{2} = \frac{50}{\sqrt{3}}$
$d_{2} = \frac{50 \sqrt{3}}{3}$
Distance between the two cars = $d_{1} - d_{2}$ = $50 \sqrt{3} - \frac{50 \sqrt{3}}{3}$
= $\frac{100 \sqrt{3}}{3}$ m = $57.73$ 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 of height 50 m. From the top of the tower, angles of depressions of two cars standing on the highway are 30∘ and 60∘. what is the distance between the cars. ?

The correct option is A 57.73mHeight of tower = 50 mLet the distance between car 1 and the tower = d1Let the distance between car 2 and the tower = d2Thus, tan30=50d1d1=50√3tan60=50d2d2=50√3d2=50√33Distance between the two cars = d1−d2 = 50√3−50√33= 100√33 m = 57.73 m

A straight highway leads to the foot of a tower of height 50 m. From the top of the tower, angles of depressions of two cars standing on the highway are $30^{\circ}$ and $60^{\circ}$ . what is the distance between the cars. ?