Question

A man standing at the top of a building observes a car at an angle of depression of 30⁰,

which is approaching the foot of the building with a uniform speed. Six seconds later the

angle of depression of the car is found to be 60⁰.Now the car is at a point 25m away from the

foot of the building.

(i) Find the height of the building

a) 25√ m b) 25 m c) 30m d)30√ m

(ii) Distance between the two positions of the car is

a) 40m b) 30m c) 60m d) 50m

(iii) Find the total time taken by the car from the starting point to reach the foot of the

building

a) 3sec b) 9sec c) 6sec d) none of these

(iv) The distance of the observer from the second position of the car

a) 100 m b) 50m c) 75m d) 30√ m​

Answer 1

Answer:

iv) in ️ABC

cos 60= BC /AC

AC = The distance of the observer from the second position of the car = 25 × 2 = 50m b)

Answer 2

Answer:

°

Step-by-step explanation:

Let AB =x m and height of building is h

In ΔPQB

tan60°=h/25

√3=h/25

h=25√3 m option ( a)

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(ii) In Δ PQA

tan 30=h/(x+25)

1/√3 = 25√3/(x+25)

x+25=√3*25√3

x+25=25*3=75

x=50 m

distance between two points  A and B

=50 m

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Time from A to B = 3 sec

So speed=50 m/6=25/3 m/s

From B to foot Q distance=25+502=75 m

so time= AQ/speed

=75/( 25/3)=3*75/25

=225/25=9 sec (b)

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(iv) The distance of the observer from the second position of the car

in Δ PBQ

BP/BQ= sec 60

BP=25*2=50 m