Question

A vertical tower of height H stands on the ground. From point A on the ground, the angle of elevation to the top is 45°. From point B, located 15 meters away from A (directly along the line extending from the base), the angle of elevation is 30°. What is the height of the tower?

• A. 15(√3-1) m
• B. 15(√3+1)/2 m
• C. 20 (√3-1) m
• D. 20(√3+1) m

Answer 1

Answer:

B. 15(√3+1)/2 m

Let the height of the tower be (H) and the distance from point A to the base of the tower be (x).

From point A:

$\tan 45^\circ=\frac{H}{x}$
Since $(\tan45^\circ=1),$

$\frac{H}{x}=1 \Rightarrow H=x$

From point B, which is 15 m farther away:

$\text{Distance from B to tower}=x+15$

The angle of elevation from B is $(30^\circ):$

$\tan30^\circ=\frac{H}{x+15}$

Since $(\tan30^\circ=\frac{1}{\sqrt3}),$

$\frac{H}{x+15}=\frac{1}{\sqrt3}$

Substitute (H=x):

$\frac{x}{x+15}=\frac{1}{\sqrt3}$

Cross multiply:

$\sqrt3x=x+15$
$x(\sqrt3-1)=15$
$x=\frac{15}{\sqrt3-1}$

Rationalize:

$x=\frac{15(\sqrt3+1)}{3-1}$
$x=\frac{15(\sqrt3+1)}{2}$
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