Question

A man on a top of a tower observes a truck at angle of depression A where tanA=1/root5 and sees that it is moving towards the base of the tower. Ten minutes later, the angle of depression of truck found to be B whre tanB=root5 if the truck is moving at uniform speed. Determine how much more time it wil take to reach the base of the tower.

Answer 1

Answer:

Hope this will help you.

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Answer 2

Time taken by truck to reach the tower C is 2.5 min  

Step-by-step explanation:

Let AB be the Vertical Tower. Suppose D and C be the positions of the truck when the angle of depression from the top of the tower is $\alpha\ and\ \beta$ respectively.

Suppose the uniform speed of the truck be $v\ mmin^-^1$

Time taken for the angle of depression to change from $\alpha\ to\ \beta$ = 10  min (Given)

$\angle EAD = \angle ADB = \alpha$ (Alternative Angles)

$\angle EAC = \angle ACB = \beta$ (Alternative Angles)

• Suppose AB = h m and BC = x m

CD = Distance covered by car in 10 min = v $mmin^{-1} \times$ 10 min = 10v m

In ΔADB,  

$tan \alpha = \frac {AB}{BD}$

So, $\frac {1}{\sqrt 5} = \frac {AB}{CD + BC} = \frac {h}{10v + x} (As\ tan \alpha = \frac {1}{\sqrt 5})$

Hence, 10v + x = $\sqrt 5h$  ""(1)

In ΔACB,  

$tan \beta = \frac {AB}{BC}$

So, $\sqrt 5 = \frac {h}{x}$

or $h = \sqrt 5 x$   """(2)

From (1) & (2) we get-

$10v + x = \sqrt 5(\sqrt 5x)$

So, 10v + x = 5x

4x = 10v

$\frac {x}{v} =\frac {10}{4} = \frac {5}{2} = 2.5$

Time Taken by truck to reach the tower from C = $\frac {x}{v}$= 2.5 min.