A man is watching from the top of a tower a boat speeding away from the tower. The boat makes an angle of depression of 45° with the man s eye when at a distance of 100 metres from the tower. After 10 seconds, the angle of depression becomes 30°. What is the approximate speed of the boat, assuming that it is running in still water?
26.28 km/hr
32.42 km/hr
24.22 km/hr
31.25 km/hr
Explanation:
Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, $\angle$ACB = 45° , $\angle$ADC = 30°, BC = 100 m
tan45°=$\dfrac{AB}{BC}$
=>1=$\dfrac{AB}{100}$
=> AB = 100 ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{100}{BD}$ (∵ Substituted the value of AB from equation 1)
=> BD =100√3
CD = (BD - BC) =(100√3−100)=100(√3−1)
It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√3−1) is covered in 10 seconds.
Required speed
=$\dfrac{Distance}{Time}=\dfrac{100(\sqrt{3}−1)}{10}$=10(1.73−1)
= 7.3 meter/seconds
= 7.3 × $\dfrac{18}{5}$ km/hr = 26.28 km/hr
Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, $\angle$ACB = 45° , $\angle$ADC = 30°, BC = 100 m
tan45°=$\dfrac{AB}{BC}$
=>1=$\dfrac{AB}{100}$
=> AB = 100 ⋯(1)
tan30°=$\dfrac{AB}{BD}$
=>$\dfrac{1}{\sqrt{3}}=\dfrac{100}{BD}$ (∵ Substituted the value of AB from equation 1)
=> BD =100√3
CD = (BD - BC) =(100√3−100)=100(√3−1)
It is given that the distance CD is covered in 10 seconds.
i.e., the distance 100(√3−1) is covered in 10 seconds.
Required speed
=$\dfrac{Distance}{Time}=\dfrac{100(\sqrt{3}−1)}{10}$=10(1.73−1)
= 7.3 meter/seconds
= 7.3 × $\dfrac{18}{5}$ km/hr = 26.28 km/hr