★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°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
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Answer:
Let BCD is a highway. A tower is standing at point D of height h. From the top of tower of point A the angle of depression is 30°. After 6 sec when car reaches at point C then angle of depression becomes 60°. Hence distance covered in 6 sec = BC From right angled ∆ADB, tan 30° = AD/BD ⇒ 1/√3 = h/BD ⇒ BD = h√3 …..(i) Again, From right angled ∆ADC, tan 60° = AD/CD ⇒ √3 = h/CD ⇒ h = √3CD …..(ii) Put the value of h in equation (i), BD = √3CD × √3 = 3CD BC + CD = 3CD 2CD = BC CD = 1/2 BC Since car is moving with uniform speed and distance CD is half of BC. Hence, time taken to cover distance CD = 1/2 × time taken to cover distance BC = 1/2 × 6 = 3 sec. Hence, required time = 3.
straight-highway-leads-foot-tower-standing-top-the-tower-observes-car-angle-depression-30
Answer:
Let BCD is a highway. A tower is standing at point D of height h. From the top of tower of point A the angle of depression is 30°. After 6 sec when car reaches at point C then angle of depression becomes 60°. Hence distance covered in 6 sec = BC From right angled ∆ADB, tan 30° = AD/BD ⇒ 1/√3 = h/BD ⇒ BD = h√3 …..(i) Again, From right angled ∆ADC, tan 60° = AD/CD ⇒ √3 = h/CD ⇒ h = √3CD …..(ii) Put the value of h in equation (i), BD = √3CD × √3 = 3CD BC + CD = 3CD 2CD = BC CD = 1/2 BC Since car is moving with uniform speed and distance CD is half of BC. Hence, time taken to cover distance CD = 1/2 × time taken to cover distance BC = 1/2 × 6 = 3 sec. Hence, required time = 3.
straight-highway-leads-foot-tower-standing-top-the-tower-observes-car-angle-depression-30
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