The angle of elevation of the top of a vertical tower from a point A, due east of it is 45°. The angle of elevation of the top of the same tower from a point B, due south of A is 30°. If the distance between A and B is $$54\sqrt{2}$$ m, then the height of the tower (in meters), is:

Let us denote the foot of the vertical tower by the point $$O$$ and its height by $$h$$ metres, so the top of the tower is the point $$T$$ with $$OT=h$$. We are told that point $$A$$ lies due east of the tower, therefore the line $$OA$$ is purely horizontal and directed east-west. Let $$OA=d$$ metres.

From point $$A$$ the angle of elevation of the top $$T$$ is $$45^\circ$$. We recall the basic trigonometric identity for a right-angled triangle:

For a right-angled triangle with opposite side $$\text{(opp)}$$, adjacent side $$\text{(adj)}$$,
$$\tan\theta=\dfrac{\text{opp}}{\text{adj}}.$$

Applying this to triangle $$\triangle OAT$$, where $$\angle OAT=45^\circ$$, the opposite side is $$OT=h$$ and the adjacent side is $$OA=d$$. Hence

$$ \tan 45^\circ=\dfrac{h}{d}. $$

But $$\tan 45^\circ=1$$, so

$$ 1=\dfrac{h}{d}\;\Longrightarrow\;h=d. $$

Thus the height of the tower equals the horizontal distance $$OA$$.

Next, the point $$B$$ is due south of $$A$$ and the distance between $$A$$ and $$B$$ is given as $$54\sqrt{2}$$ m. Since the direction is due south, the line $$AB$$ is purely north-south, so

$$ AB = 54\sqrt{2}\ \text{m}. $$

Let us set $$AB = s$$. Hence $$s = 54\sqrt{2}$$.

Because $$B$$ is directly south of $$A$$, the horizontal coordinates can be represented on a right-angled grid as follows:

• The tower foot $$O$$ is at the origin $$(0,0)$$.
• Point $$A$$ is $$d$$ metres to the east: $$(d,0)$$.
• Point $$B$$ is $$s$$ metres south of $$A$$: $$(d,-s)$$.

Therefore, the horizontal ground distance from $$B$$ to the tower foot $$O$$ is obtained from the Pythagorean theorem:

$$ OB=\sqrt{(d-0)^2+(-s-0)^2}=\sqrt{d^2+s^2}. $$

Now, from point $$B$$ the angle of elevation of the top $$T$$ is $$30^\circ$$. Using the same tangent formula in triangle $$\triangle OBT$$, we have

$$ \tan 30^\circ = \dfrac{\text{opposite side }(OT=h)}{\text{adjacent side }(OB=\sqrt{d^2+s^2})}. $$

Because $$\tan 30^\circ = \dfrac{1}{\sqrt{3}}$$, the equation reads

$$ \dfrac{1}{\sqrt{3}} = \dfrac{h}{\sqrt{d^2+s^2}}. $$

Cross-multiplying gives

$$ h = \dfrac{1}{\sqrt{3}}\;\sqrt{d^2+s^2}. $$

But we have already shown that $$h=d$$, therefore we substitute $$h$$ by $$d$$:

$$ d = \dfrac{1}{\sqrt{3}}\;\sqrt{d^2+s^2}. $$

To eliminate the square root, we square both sides:

$$ d^2 = \dfrac{1}{3}\,(d^2+s^2). $$

Multiplying by $$3$$ yields

$$ 3d^2 = d^2 + s^2. $$

Rearranging,

$$ 3d^2 - d^2 = s^2 \;\Longrightarrow\; 2d^2 = s^2. $$

Hence

$$ d^2 = \dfrac{s^2}{2}\quad\Longrightarrow\quad d = \dfrac{s}{\sqrt{2}}. $$

We know $$s = 54\sqrt{2}$$, so substituting this value gives

$$ d = \dfrac{54\sqrt{2}}{\sqrt{2}} = 54\ \text{m}. $$

Finally, since $$h=d$$, we have

$$ h = 54\ \text{m}. $$

Hence, the correct answer is Option D.