Tower Height Calculation Using Angles

Let the height of the vertical tower be $h$ meters. Let the base of the tower be point $O$ and the top be point $P$. So, $OP = h$.

A person stands at point $A$ on the horizontal ground, due north of the tower. The angle of elevation of $P$ from $A$ is $45^\circ$. In the right-angled triangle $\triangle PAO$, we have:

• $\tan(\angle PAO) = \frac{OP}{OA}$
• $\tan(45^\circ) = \frac{h}{OA}$
• $1 = \frac{h}{OA}$
• $OA = h$

Another person stands at point $B$, which is $50$ m west of $A$. The angle of elevation of $P$ from $B$ is $30^\circ$. In the right-angled triangle $\triangle PBO$, we have:

• $\tan(\angle PBO) = \frac{OP}{OB}$
• $\tan(30^\circ) = \frac{h}{OB}$
• $\frac{1}{\sqrt{3}} = \frac{h}{OB}$
• $OB = h\sqrt{3}$

Ground Geometry and Pythagorean Theorem

Point $A$ is due North of the tower base $O$. Point $B$ is $50$ m West of $A$. This means the angle $\angle OAB$ is a right angle ($90^\circ$) because North and West directions are perpendicular. We can apply the Pythagorean theorem to the right-angled triangle $\triangle OAB$ on the ground:

• $OB^2 = OA^2 + AB^2$
• We know $OA = h$ and $AB = 50$ m.
• $OB^2 = h^2 + 50^2$
• $OB^2 = h^2 + 2500$

Solving for Height

Now, substitute the expression for $OB$ from the trigonometry step ($OB = h\sqrt{3}$) into the Pythagorean equation:

• $(h\sqrt{3})^2 = h^2 + 2500$
• $3h^2 = h^2 + 2500$
• $3h^2 - h^2 = 2500$
• $2h^2 = 2500$
• $h^2 = \frac{2500}{2}$
• $h^2 = 1250$
• $h = \sqrt{1250}$
• $h = \sqrt{625 \times 2}$
• $h = 25\sqrt{2}$

The height of the tower is $25\sqrt{2}$ meters.