35
Tejas is standing at the top of a building and observes a car at an angle of depression of $30^\circ$ as it approaches the base of the building at a uniform speed. $6$ seconds later, the angle of depression increases to $60^\circ$, and at that moment, the car is $25$ m away from the building.
Based on the information given above, answer the following questions :
(i) What is the height of the building?
(ii) What is the distance between the two positions of the car?
(iii) (a) What would be the total time taken by the car to reach the foot of the building from the starting point?
OR
(iii) (b) What is the distance of the observer from the car when it makes an angle of $60^\circ$?
Based on the information given above, answer the following questions :
(i) What is the height of the building?
(ii) What is the distance between the two positions of the car?
(iii) (a) What would be the total time taken by the car to reach the foot of the building from the starting point?
OR
(iii) (b) What is the distance of the observer from the car when it makes an angle of $60^\circ$?
Show SolutionHide Solution↓
(i) In $\triangle ABC$,
$\tan 60^\circ = \sqrt{3} = \frac{AB}{25}$ (1/2 Mark)
$\Rightarrow AB = 25\sqrt{3}$ (1/2 Mark)
$\therefore$ Height of building = $25\sqrt{3}$ m
(ii) In $\triangle ABD$,
$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD}$ (1/2 Mark)
$\Rightarrow BD = 75$
$\therefore$ Distance between two positions of car = $75 - 25 = 50$ m (1/2 Mark)
(iii) (a) Time taken to cover the distance of $50$ m = $6$ sec (1 Mark)
$\therefore$ Time taken to cover the distance of $75$ m = $\frac{6}{50} \times 75$ (1 Mark)
$= 9$ sec
OR
(iii) (b) In $\triangle ABC$,
$\cos 60^\circ = \frac{BC}{AC}$ (1 Mark)
$\Rightarrow \frac{1}{2} = \frac{25}{AC}$
$\Rightarrow AC = 50$ (1 Mark)
$\therefore$ Distance of the observer from car when it makes the angle of $60^\circ = 50$ m
$\tan 60^\circ = \sqrt{3} = \frac{AB}{25}$ (1/2 Mark)
$\Rightarrow AB = 25\sqrt{3}$ (1/2 Mark)
$\therefore$ Height of building = $25\sqrt{3}$ m
(ii) In $\triangle ABD$,
$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD}$ (1/2 Mark)
$\Rightarrow BD = 75$
$\therefore$ Distance between two positions of car = $75 - 25 = 50$ m (1/2 Mark)
(iii) (a) Time taken to cover the distance of $50$ m = $6$ sec (1 Mark)
$\therefore$ Time taken to cover the distance of $75$ m = $\frac{6}{50} \times 75$ (1 Mark)
$= 9$ sec
OR
(iii) (b) In $\triangle ABC$,
$\cos 60^\circ = \frac{BC}{AC}$ (1 Mark)
$\Rightarrow \frac{1}{2} = \frac{25}{AC}$
$\Rightarrow AC = 50$ (1 Mark)
$\therefore$ Distance of the observer from car when it makes the angle of $60^\circ = 50$ m