Tower Height Calculation Using Angles

This problem involves finding the height of a tower using trigonometry based on the angles of depression observed from the top of the tower to two cars on the ground.

Let the height of the tower be denoted by $h$. Let the distances of the two cars from the foot of the tower be $d_1$ and $d_2$. The angles of depression are given as $\alpha$ and $\beta$, with $\alpha > \beta$. The angle of depression from the observer to an object is equal to the angle of elevation from the object to the observer.

Since $\alpha > \beta$, the car corresponding to angle $\alpha$ is closer to the tower. Let $d_1$ be the distance of this closer car and $d_2$ be the distance of the farther car.

We are given:

• Distance between the two cars = 120 m. So, $d_2 = d_1 + 120$ m.
• $\tan \alpha = \sqrt{3}$
• $\tan \beta = \frac{1}{\sqrt{3}}$

Solving for Tower Height

We can set up trigonometric relationships using the tangent function for the two right-angled triangles formed:

1. For the closer car (angle $\alpha$):

   $\tan \alpha = \frac{\text{Height of tower}}{\text{Distance of closer car}} = \frac{h}{d_1}$

   Substituting the given value:

   $\sqrt{3} = \frac{h}{d_1} \implies d_1 = \frac{h}{\sqrt{3}}$

2. For the farther car (angle $\beta$):

   $\tan \beta = \frac{\text{Height of tower}}{\text{Distance of farther car}} = \frac{h}{d_2}$

   Substituting the given value:

   $\frac{1}{\sqrt{3}} = \frac{h}{d_2} \implies d_2 = h\sqrt{3}$

3. Now, use the relationship between the distances $d_1$ and $d_2$: $d_2 = d_1 + 120$.

   Substitute the expressions for $d_1$ and $d_2$ in terms of $h$:

   $h\sqrt{3} = \frac{h}{\sqrt{3}} + 120$

4. Solve the equation for $h$:

   $h\sqrt{3} - \frac{h}{\sqrt{3}} = 120$

   Combine the terms on the left side:

   $h \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) = 120$

   $h \left( \frac{3 - 1}{\sqrt{3}} \right) = 120$

   $h \left( \frac{2}{\sqrt{3}} \right) = 120$

   Isolate $h$:

   $h = 120 \times \frac{\sqrt{3}}{2}$

   $h = 60\sqrt{3}$ m

The height of the tower is $60\sqrt{3}$ m.