Question

Question :-

A straight highway leads to the foot of a tower a man standing at the top of the tower observes a car at an angle of depression of 30 degree which is approaching the foot of the tower with the uniform speed 6 seconds later the angle of depression of the car is found to be 60 degree find the time taken by the car to reach the foot of the tower from this point .

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Answer 1

Question ⤵

A straight highway leads to the foot of a tower a man standing at the top of the tower observes a car at an angle of depression of 30 degree which is approaching the foot of the tower with the uniform speed 6 seconds later the angle of depression of the car is found to be 60 degree find the time taken by the car to reach the foot of the tower from this point .

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Answer 2

Answer:

(Please refer to the picture given):

Let the distance of CD be 'x' meter.

Speed= Distance/Time

=x/6 m/s.

Since the speed is uniform throughout the journey, the same speed will be used to cover BD as well......(1)

Let the tower AB have a height of 'h' meter.

Now, tan60=h/BD

tan60=√3

This implies that h/BD=√3

Which means that BD=h/√3........(2)

tan30=h/BC

tan30=1/√3

This implies that h/BC=1/√3

Which means that BC=h√3..........(3)

BC-BD=x ( According to the figure)

This implies that

h√3-h/√3=x (substituting the values of BC and BD from (2) and (3)).

This implies

(3h-h)/√3= 2h/√3=x

This implies that h= x√3/2........(4)

Now, BD=h/√3 (from(2))

Substituting the value of h from (4), we get that:

BD= (x√3/2)÷2=x/2............(5)

Now, Speed= Distance/time.

We found the distance of BD to be x/2 from equation (5). And Speed will be x/6 from equation (1)

This implies that

x/6=(x/2)÷time taken

This implies that time taken is (x/2)÷(x/6)= 3 seconds.

Hence, the time taken to cover that Distance is 3 seconds