Question

$ABC$ is a triangular park with $AB = AC = 100$ m. A TV tower stands at the mid-point of $BC$. The angles of elevation of the top of the tower at $A, B, C$ are $45^\circ, 60^\circ, 60^\circ$ respectively. The height of the tower is

• A. 50 m
• B. $50\sqrt{3}$ m
• C. $50\sqrt{2}$ m
• D. $50(3 - \sqrt{3})$m

Answer 1

Answer: (B)

$50\sqrt{3}$ m

Answer 2

Let P be the midpoint of BC, where the TV tower stands. Let Q be the top of the tower. The height of the tower is $h = PQ$.
The park is a triangle ABC with $AB = AC = 100$ m. P is the midpoint of BC.
The angles of elevation of the top of the tower Q from A, B, C are $\angle QAP = 45^\circ$, $\angle QBP = 60^\circ$, and $\angle QCP = 60^\circ$.
Since the tower is vertical, $\triangle QAP$, $\triangle QBP$, and $\triangle QCP$ are right-angled triangles at P.

In $\triangle QAP$:
$\tan(\angle QAP) = \frac{PQ}{AP}$
$\tan(45^\circ) = \frac{h}{AP}$
$1 = \frac{h}{AP} \implies AP = h$.

In $\triangle QBP$:
$\tan(\angle QBP) = \frac{PQ}{BP}$
$\tan(60^\circ) = \frac{h}{BP}$
$\sqrt{3} = \frac{h}{BP} \implies BP = \frac{h}{\sqrt{3}}$.

In $\triangle QCP$:
$\tan(\angle QCP) = \frac{PQ}{CP}$
$\tan(60^\circ) = \frac{h}{CP}$
$\sqrt{3} = \frac{h}{CP} \implies CP = \frac{h}{\sqrt{3}}$.
Since P is the midpoint of BC, $BP = CP$, which is consistent with the angles of elevation from B and C being equal.

Triangle ABC is an isosceles triangle with $AB = AC$. Since P is the midpoint of BC, AP is the median to the base BC. In an isosceles triangle, the median to the base is also the altitude to the base. Therefore, AP is perpendicular to BC, and $\triangle APB$ is a right-angled triangle with the right angle at P.

Apply the Pythagorean theorem in $\triangle APB$:
$AB^2 = AP^2 + BP^2$
Substitute the given value $AB = 100$ m and the expressions for AP and BP in terms of h:
$100^2 = h^2 + \left(\frac{h}{\sqrt{3}}\right)^2$
$10000 = h^2 + \frac{h^2}{3}$
$10000 = h^2 \left(1 + \frac{1}{3}\right)$
$10000 = h^2 \left(\frac{4}{3}\right)$
$h^2 = 10000 \times \frac{3}{4}$
$h^2 = 2500 \times 3$
$h^2 = 7500$
$h = \sqrt{7500}$
$h = \sqrt{2500 \times 3}$
$h = 50\sqrt{3}$ m.

The height of the tower is $50\sqrt{3}$ m.