Question

A pilot in an aeroplane observes that Vashi bridge is one side of a plane and Warli sea-link is on the opposite side. The angle of depression of Vashi bridge and Warli sea-link are $${60}^{\circ }$$ and $${30}^{\circ }$$ respectively. If the aeroplanes is at a height of 5500$$\sqrt{3}$$m at that time. What is the distance between Vashi bridge and Warli sea-link?

Answer 1

Hint:- We can take variable x and y as the horizontal distance between the aeroplane and Vashi bridge and Warli sea-link. And then we can use trigonometric identity like $$\tan \theta ={\displaystyle \frac{\text{Perpendicular}}{\text{Base}}}$$ to find the value of x and y as the angle of depression is given as $${60}^{\circ }$$ and $${30}^{\circ }$$.

Complete step-by-step answer:
Let us draw the diagram which will make it easy to solve the problem.

Now as seen from the above diagram Aeroplane is at point A, Vashi bridge is at point B and Warli sea-link is at point C.
So, now we have to find the value of x and y and then the distance between Vashi bridge and Warli sea link will be equal to x + y.
So, now as we know that $$\tan \theta ={\displaystyle \frac{\text{Perpendicular}}{\text{Base}}}$$ .
So, in triangle ABD, $$\tan {60}^{\circ }={\displaystyle \frac{AD}{BD}}$$
As we know that $$\tan {60}^{\circ }=\sqrt{3}$$
So, $$\sqrt{3}={\displaystyle \frac{AD}{BD}}$$
Now cross multiplying the above equation.
$$BD={\displaystyle \frac{AD}{\sqrt{3}}}={\displaystyle \frac{5500\sqrt{3}}{\sqrt{3}}}=5500$$m
So, x = 5500 metres
In triangle ADC, $$\tan {30}^{\circ }={\displaystyle \frac{AD}{DC}}$$ and $$\tan {30}^{\circ }={\displaystyle \frac{1}{\sqrt{3}}}$$
So, $${\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{AD}{DC}}$$
Now cross multiplying the above equation.
$$DC=AD\sqrt{3}=\left(5500\sqrt{3}\right)\sqrt{3}=5500\times 3=16500$$m
So, y = 16500 metres
Hence, the distance between Vashi bridge and Warli sea-link will be x + y = 5500 + 16500 = 22000 metres.

Note:- Whenever we come up with this type of problem then there is a trick to find the length of any side (here base) of the triangle if the length of any one side (here height) and one angle is given. Like if the height is given and asked the base length (distance between Vashi bridge and Warli sea-link) then we had to use the identity of $$\tan \theta$$ or $$\cot \theta$$ because these two identities involve base length and height. And if the hypotenuse of the triangle is given or asked then we have to use $$\sin \theta$$ or $$\cos \theta$$ because these two involve hypotenuses.