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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 $60^\circ$?
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(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 (1 Mark)
(ii) In $\triangle ABD$,
$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD}$ (1/2 Mark)
$\Rightarrow BD = 75$ (1/2 Mark)
$\therefore$ Distance between two positions of car = $75 - 25 = 50$ m (1 Mark)
(iii) (a) Time taken to cover the distance of 50 m = 6 sec (1 Mark)
Time taken to cover the distance of 75 m = $\frac{6}{50} \times 75$
$= 9 \text{ sec}$ (1 Mark)
OR
(iii) (b) In $\triangle ABC$,
$\cos 60^\circ = \frac{BC}{AC}$ (1 Mark)
$\Rightarrow \frac{1}{2} = \frac{25}{AC}$
AC = 50 (1 Mark)
$\therefore$ Distance of the observer from car when it makes the angle of $60^\circ = 50 \text{ m}$
$\tan 60^\circ = \sqrt{3} = \frac{AB}{25}$ (1/2 Mark)
$\Rightarrow AB = 25\sqrt{3}$ (1/2 Mark)
$\therefore$ Height of building = $25\sqrt{3}$ m (1 Mark)
(ii) In $\triangle ABD$,
$\tan 30^\circ = \frac{1}{\sqrt{3}} = \frac{25\sqrt{3}}{BD}$ (1/2 Mark)
$\Rightarrow BD = 75$ (1/2 Mark)
$\therefore$ Distance between two positions of car = $75 - 25 = 50$ m (1 Mark)
(iii) (a) Time taken to cover the distance of 50 m = 6 sec (1 Mark)
Time taken to cover the distance of 75 m = $\frac{6}{50} \times 75$
$= 9 \text{ sec}$ (1 Mark)
OR
(iii) (b) In $\triangle ABC$,
$\cos 60^\circ = \frac{BC}{AC}$ (1 Mark)
$\Rightarrow \frac{1}{2} = \frac{25}{AC}$
AC = 50 (1 Mark)
$\therefore$ Distance of the observer from car when it makes the angle of $60^\circ = 50 \text{ m}$