Question

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?

Answer 1

Let AB be the tower and C and D be the positions of the boat.

then∠ACB=45∘,∠ADB=30∘,AC=100m

Let=>AB=h =AB/AC=tan45=1

=>AB=AC=100m

and,AB/AD=tan30=1/√3

=>AD=AB∗3–√3=100√3m

=>CD=AD−AC=100√3−100

=>CD=100(√3−1)m

We Know Speed = Distance/Time

=>Speed=[100(√3−1)/10m/sec=7.32

Please note answer we need in Km/Hr =>Speed=7.32∗18/5Km/Hr

=>Speed=26.352 Km/Hr  ANS

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Answer 2

Answer:

The approximate speed of the boat, assuming that it is running in still water is 7.3205 m/sec.

Step-by-step explanation:

Refer the attached figure :

The boat makes an angle of depression of 45° with the man's eye when at a distance of 100 meters from the tower i.e. ∠ACB=45° and BC=100 m

After 10 seconds, the angle of depression becomes 30°. i.e. ∠ADB=30°

Let CD be x

So, BD = BC+CD=100+x

In ΔABC

$tan \theta = \frac{Perpendicular}{Base}$

$tan 45^{\circ} = \frac{AB}{BC}$

$1 = \frac{AB}{100}$

$100 =AB$

In ΔABD

$tan \theta = \frac{Perpendicular}{Base}$

$tan 30^{\circ} = \frac{AB}{BD}$

$\frac{1}{\sqrt{3}} = \frac{AB}{100+x}$

$\frac{1}{\sqrt{3}} = \frac{100}{100+x}$

$100+x= \frac{100}{\frac{1}{\sqrt{3}}}$

$100+x= 173.205$

$x= 173.205-100$

$x= 73.205$

So, Distance traveled in 10 seconds = 73.205 m

Speed of boat = $\frac{Distance}{Time}$

Speed of boat = $\frac{73.205}{10}$

Speed of boat = $7.3205$

Hence the approximate speed of the boat, assuming that it is running in still water is 7.3205 m/sec.