Question

A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 min for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower? Options: a. 5( √3+ 1 ) b. 6 (√3 +√2) c. 7 (√3- 1) d. 8 (√3-2)

Answer 1

Answer:

The correct answer is option a.

a. 5( √3+ 1 )

Step-by-step explanation:

Given that the angle changes from 45° to 60° in 10 minutes.

This situation can be represented as right angled triangles $\triangle$ABC (in the starting when angle is 45°)and $\triangle$ABD (after 10 minutes when the angle is 60°).

AB is the tower (A be its top and B be its base).

Now, we need to find the time to be taken to cover the distance D to B.

First of all, let us consider $\triangle$ABC.

Using tangent property:

$tan\theta =\dfrac{Perpendicular}{Base}\\\Rightarrow tan 45=\dfrac{AB}{BC}\\\Rightarrow 1=\dfrac{h}{BC}\\\Rightarrow h = BC$

Using tangent property in $\triangle$ABD:

$\Rightarrow tan 60=\dfrac{AB}{BD}\\\Rightarrow \sqrt3=\dfrac{h}{BD}\\\Rightarrow BD = \dfrac{h}{ \sqrt3}\ units$

Now distance traveled in 10 minutes, CD  = BC - BD

$\Rightarrow h - \dfrac{h}{\sqrt3}\\\Rightarrow \dfrac{(\sqrt3-1)h}{\sqrt3}$

$Speed =\dfrac{Distance }{Time}$

$\Rightarrow \dfrac{(\sqrt3-1)h}{10\sqrt3}$

Now, we can say that more distance to be traveled to reach the base of tower is BD i.e. '$\bold{\dfrac{h}{\sqrt3}}$'

So, more time required = Distance left divided by Speed

$\Rightarrow \dfrac{\dfrac{h}{\sqrt3}}{\dfrac{(\sqrt3-1)h}{10\sqrt3}}\\\Rightarrow \dfrac{h\times 10\sqrt3}{\sqrt3(\sqrt3-1)h}\\\Rightarrow \dfrac{10 (\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)} (\text{Rationalizing the denominator})\\\Rightarrow \dfrac{10 (\sqrt3+1)}{3-1}\\\Rightarrow \dfrac{10 (\sqrt3+1)}{2}\\\Rightarrow 5(\sqrt3+1)}$

So, The correct answer is option a.

a. 5( √3+ 1 )