The angles of depression of two houses of the same height from the top of a building are 45° and 30° towards the east. If the two houses are 50 m apart, what will be the height of the building in metres?

Correct option is B

Given:

Angles of depression to two identical houses (same height) from a building: $$45^\circ$$​and $$30^\circ$$​​

Distance between the two houses (on the ground) = 50 m

Formula Used:

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

​$$\tan 45^o = 1$$​​

​$$\tan 30^o = \frac1{\sqrt3}$$​​

Solution:

Let height of the building be h
Let distance of nearer house from the building be x  

Then, from triangle with angle $$45^\circ:$$​​

​$$\tan45^\circ = \frac{h}{x} \\ \ \\ 1 = \frac{h}{x} \\ \ \\ h = x$$​​

From triangle with angle $$30^\circ$$​:

​$$\tan30^\circ = \frac{h}{x + 50} \\ \ \\ \frac{1}{\sqrt{3}} = \frac{h}{x + 50} \\ \ \\ h = \frac{x + 50}{\sqrt{3}}$$​​

Now,

​$$\frac{x + 50}{\sqrt{3}} = xx + 50 = x\sqrt{3} \\ \ \\ x\sqrt{3} - x = 50 \\ \ \\ x(\sqrt{3} - 1) = 50 \\ \ \\ x = \frac{50}{\sqrt{3} - 1}\\ \ \\x = 25(\sqrt{3} + 1) \ m$$

So, h = $$25(\sqrt{3} + 1) \ m$$​​