(D) A boy at point 'B', standing at a distance of 12 metres from the f

1.	(D) A boy at point 'B', standing at a distance of 12 metres from the foot of a tower 'MS' observes aman on the top of the tower at an angle of elevation of 30°. The man on the top of the toweobserves a stationary bus at point 'C' on the ground at an angle of depression of 60° as shown inthe figure. Find the distance of the boy from the bus. (V3 = 1.73)
Answer» It is given that the angle of depression is30°30°and after six seconds, the angle of depression is60°60°. Let AB be the height of the tower,C is the initial position of the car and D is the position after six seconds. InΔADBΔADB, ABDB=tan60°ABDB=√3DB=AB√3ABDB=tan60°ABDB=3DB=AB3 InΔABCΔABC, ABBC=tan30°ABBD+DC=1√3AB√3=BD+DCABBC=tan30°ABBD+DC=13AB3=BD+DC Simplify further, AB√3−AB√3=DCDC=3AB−AB√3=2AB√3AB3−AB3=DCDC=3AB−AB3=2AB3 The speed of the car is, Speed=DistanceTime=2AB√36=AB3√3m/sSpeed=DistanceTime=2AB36=AB33 m/s The time taken to travel CD that is,2AB√32AB3distance is66seconds. The time taken to travel BD that is,√33distance is, Time=62AB√3×AB√3=62=3secondsTime=62AB3×AB3=62=3 seconds Therefore, car will take3seconds3 secondsto reach the foot of the car. 516 m is to a distance between the bus 🚌 and the boy

(D) A boy at point 'B', standing at a distance of 12 metres from the foot of a tower 'MS' observes aman on the top of the tower at an angle of elevation of 30°. The man on the top of the toweobserves a stationary bus at point 'C' on the ground at an angle of depression of 60° as shown inthe figure. Find the distance of the boy from the bus. (V3 = 1.73)

Answer»

It is given that the angle of depression is30°30°and after six seconds, the angle of depression is60°60°.

Let AB be the height of the tower,C is the initial position of the car and D is the position after six seconds.

InΔADBΔADB,

ABDB=tan60°ABDB=√3DB=AB√3ABDB=tan60°ABDB=3DB=AB3

InΔABCΔABC,

ABBC=tan30°ABBD+DC=1√3AB√3=BD+DCABBC=tan30°ABBD+DC=13AB3=BD+DC

Simplify further,

AB√3−AB√3=DCDC=3AB−AB√3=2AB√3AB3−AB3=DCDC=3AB−AB3=2AB3

The speed of the car is,

Speed=DistanceTime=2AB√36=AB3√3m/sSpeed=DistanceTime=2AB36=AB33 m/s

The time taken to travel CD that is,2AB√32AB3distance is66seconds.

The time taken to travel BD that is,√33distance is,

Time=62AB√3×AB√3=62=3secondsTime=62AB3×AB3=62=3 seconds

Therefore, car will take3seconds3 secondsto reach the foot of the car.

516 m is to a distance between the bus 🚌 and the boy