Suppose a person is standing on a tower of height 15(√3+1) m and observing a car coming towards the tower. He observed that angle of depression changes from 30° to 45°, in 3 seconds. Find the speed of the car.
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)
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[FIGURE IS IN THE ATTACHMENT]
Given:
Height of the tower( PQ)= 15(√3+1) m
∠PQA= 30°, ∠PBQ= 45°
Time= 3 secs
Let the Distance of AB= y m
Distance of BQ= x m
In ∆ PQB
tan 45° = PQ/BQ
1= 15(√3+1)/ X
x= 15(√3+1) m………….(1)
In ∆ PQA
tan 30° = PQ/AQ = PQ/AB+BQ
1/√3 = 15(√3+1)/(x+y)
x+y = 15√3(√3+1)...........(2)
Put the value of x from eq 1 in eq 2
x+y = 15√3(√3+1)
15(√3+1)+ y = 15√3(√3+1)
y= 15√3(√3+1) - 15(√3+1)
y= 15(√3+1)( √3-1)
y= 15 (√3²- 1²)
y= 15 (3-1)
y= 15×2
y= 30 m
The car is moving from A to B in 3 seconds.
Speed= Distance/time
Speed= 30/3
Speed= 10 m/s
Hence, the speed of the car is 10 m/s.
HOPE THIS WILL HELP YOU...
Given:
Height of the tower( PQ)= 15(√3+1) m
∠PQA= 30°, ∠PBQ= 45°
Time= 3 secs
Let the Distance of AB= y m
Distance of BQ= x m
In ∆ PQB
tan 45° = PQ/BQ
1= 15(√3+1)/ X
x= 15(√3+1) m………….(1)
In ∆ PQA
tan 30° = PQ/AQ = PQ/AB+BQ
1/√3 = 15(√3+1)/(x+y)
x+y = 15√3(√3+1)...........(2)
Put the value of x from eq 1 in eq 2
x+y = 15√3(√3+1)
15(√3+1)+ y = 15√3(√3+1)
y= 15√3(√3+1) - 15(√3+1)
y= 15(√3+1)( √3-1)
y= 15 (√3²- 1²)
y= 15 (3-1)
y= 15×2
y= 30 m
The car is moving from A to B in 3 seconds.
Speed= Distance/time
Speed= 30/3
Speed= 10 m/s
Hence, the speed of the car is 10 m/s.
HOPE THIS WILL HELP YOU...
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