JEE Mathematics 2023 Question with Solution

Given: Height of the tower $AB = 5 \, \text{m}$. The angle of elevation from one person is $45^\circ$ and from the other person is $30^\circ$.

Find: The distance between the two persons, that is $PQ$.

In $\triangle ABP$,

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

Since $\tan 45^\circ = 1$,

$$1 = \frac{5}{AP}$$

So,

$$AP = 5 \, \text{m}$$

In $\triangle ABQ$,

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

Using $\tan 30^\circ = \frac{1}{\sqrt{3}}$,

$$\frac{1}{\sqrt{3}} = \frac{5}{AQ}$$

Hence,

$$AQ = 5\sqrt{3} \, \text{m}$$

Since one person is due South and the other is due West of the tower, $AP \perp AQ$. Therefore, in $\triangle APQ$,

$$PQ^2 = AP^2 + AQ^2$$

Substituting the values,

$$PQ^2 = 5^2 + (5\sqrt{3})^2$$ $$PQ^2 = 25 + 75 = 100$$

Thus,

$$PQ = \sqrt{100} = 10 \, \text{m}$$

Therefore, the distance between the two persons is $10 \, \text{m}$. The correct option is A. The solution labels option B, but the working clearly gives $10$, which matches option A.

Given: Tower height $= 5 \, \text{m}$, one observer is due South and the other is due West.

Find: Distance between the two observers.

Let the foot of the tower be at the origin. Place the South observer at distance $AP$ along one axis and the West observer at distance $AQ$ along the perpendicular axis.

From the first observer,

$$\tan 45^\circ = \frac{5}{AP}$$

so

$$AP = 5$$

From the second observer,

$$\tan 30^\circ = \frac{5}{AQ}$$

so

$$AQ = 5\sqrt{3}$$

The observers lie on perpendicular directions from the tower, so their separation is the diagonal of a right triangle:

$$PQ = \sqrt{AP^2 + AQ^2}$$ $$= \sqrt{5^2 + (5\sqrt{3})^2}$$ $$= \sqrt{25 + 75}$$ $$= \sqrt{100} = 10$$

Therefore, the required distance is $10$.