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A man on the top of a vertical observation tower observers a car moving at a uniform speed coming directly towards it. If it takes 8 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
8 min 17 second
10 min 57 second
14 min 34 second
12 min 23 second
Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car. Then, angleADC = 30° , angleACB = 45° Let AB = h, BC = x, CD = y tan45°=AB/BC=h/x =>1=h/x =>h=x?(1) tan30°=AB/BD=AB/(BC + CD)=h/(x+y) =>1/√3=h/(x+y) =>x + y = √3h =>y = √3h - x =>y = √3h−h(? Substituted the value of x from equation 1 ) =>y = h(√3−1) Given that distance y is covered in 8 minutes. i.e, distance h(√3−1)is covered in 8 minutes. Time to travel distance x = Time to travel distance h (? Since x = h as per equation 1). Let distance h is covered in t minutes. since distance is proportional to the time when the speed is constant, we have h(√3−1)∝8?(A) h∝t?(B) (A)/(B)=>h(√3−1)/h=8/t =>(√3−1)=8/t =>t=8/(√3−1)=8/(1.73−1) =8/.73=800/73minutes =10 70/73minutes ≈10minutes 57seconds
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