PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of the top of the tower at P, Q and R are respectively 45$$^\circ$$, 30$$^\circ$$ and 30$$^\circ$$, then the height of the tower (in m) is:

Let the points Q and R lie on the horizontal ground along the X-axis, and put their coordinates as $$Q(-a,0)$$ and $$R(a,0)$$. Because the tower is erected at the mid-point of QR, the foot of the tower is $$M(0,0)$$ and the top of the tower is $$T(0,h)$$, where $$h$$ is the required height.

Since the park is isosceles with $$PQ = PR = 200\ \text{m}$$, place the vertex P somewhere above the X-axis at $$P(0,y)$$. (We shall soon see that this choice is consistent.)

First, from the given equalities $$PQ = PR$$ we write the two distance formulas:

$$PQ^2 = (0 + a)^2 + (y - 0)^2 = a^2 + y^2,$$

$$PR^2 = (0 - a)^2 + (y - 0)^2 = a^2 + y^2.$$

Both must equal $$200^2$$, so we have the single condition

$$a^2 + y^2 = 200^2. \qquad (1)$$

Now use the data about the angles of elevation.

At Q the horizontal distance to the tower’s foot M is $$QM = a$$. The angle of elevation there is $$30^\circ$$, so, stating the tangent formula $$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}},$$ we have

$$\tan 30^\circ = \frac{h}{a} \;\;\Longrightarrow\;\; \frac{1}{\sqrt3} = \frac{h}{a} \;\;\Longrightarrow\;\; h = \frac{a}{\sqrt3}. \qquad (2)$$

The same angle $$30^\circ$$ is seen from R, and because $$RM = a$$, relation (2) is confirmed.

From P the horizontal ground distance to M is $$PM = y$$, and the angle of elevation is $$45^\circ$$. Using $$\tan 45^\circ = 1$$ gives

$$\tan 45^\circ = \frac{h}{y} \;\;\Longrightarrow\;\; 1 = \frac{h}{y} \;\;\Longrightarrow\;\; h = y. \qquad (3)$$

Equate the two expressions (2) and (3) for the same height h:

$$y = \frac{a}{\sqrt3}. \qquad (4)$$

Substitute (4) into the distance condition (1):

$$a^2 + \left(\frac{a}{\sqrt3}\right)^2 = 200^2 \;\;\Longrightarrow\;\; a^2 + \frac{a^2}{3} = 200^2.$$

Combine like terms:

$$\frac{4a^2}{3} = 200^2 \;\;\Longrightarrow\;\; a^2 = \frac{3}{4}\, 200^2.$$

Compute $$a$$:

$$a^2 = \frac{3}{4}\times 40000 = 30000 \;\;\Longrightarrow\;\; a = \sqrt{30000} = 100\sqrt3.$$

Finally, use (2) or (3) to get $$h$$. Using (2):

$$h = \frac{a}{\sqrt3} = \frac{100\sqrt3}{\sqrt3} = 100.$$

Thus the tower is $$100 \text{ m}$$ high.

Hence, the correct answer is Option B.