Solving Tower Width Using Angles of Depression and Elevation

This solution calculates the width of the road between two towers ($T_1$ and $T_2$) using the given angles of depression and elevation.

Problem Setup

Let:

• Height of tower $T_1$, $h_1 = 60$ m.
• Height of tower $T_2$, $h_2 = 80$ m.
• Width of the road between the towers, $d$.
• Angle of depression from the top of $T_1$ to the foot of $T_2$, $\alpha$.
• Angle of elevation from the top of $T_1$ to the top of $T_2$, $\beta$.

Given condition: $\alpha = 2\beta$.

Diagram and Trigonometric Relations

Consider the scenario: Tower $T_1$ has top C and base A. Tower $T_2$ has top D and base B. The road width is $AB = d$. $AC = 60$ m, $BD = 80$ m. $AC \perp AB$ and $BD \perp AB$.

Draw a horizontal line from C, meeting BD at E. Then $CE = AB = d$, $BE = AC = 60$ m, and $ED = BD - BE = 80 - 60 = 20$ m.

We establish the trigonometric relationships:

• Angle of Elevation ($\beta$): In the right triangle $\triangle CED$ ($\angle CED = 90^\circ$), the angle of elevation from C to D is $\beta$.

  $ \tan(\beta) = \frac{ED}{CE} = \frac{20}{d} $

• Angle of Depression ($\alpha$): The angle of depression from C to the foot of $T_2$ (point B) is $\alpha$. This angle relates the vertical height $AC$ and the horizontal distance $AB$.

  $ \tan(\alpha) = \frac{AC}{AB} = \frac{60}{d} $

Calculation

Using the given relation $\alpha = 2\beta$, we have $\tan(\alpha) = \tan(2\beta)$.

Apply the tangent double angle identity: $\tan(2\beta) = \frac{2\tan(\beta)}{1 - \tan^2(\beta)}$

Substitute the trigonometric values:

$ \frac{60}{d} = \frac{2 \times \left(\frac{20}{d}\right)}{1 - \left(\frac{20}{d}\right)^2} $

Simplify the expression:

$ \frac{60}{d} = \frac{\frac{40}{d}}{1 - \frac{400}{d^2}} $

$ \frac{60}{d} = \frac{\frac{40}{d}}{\frac{d^2 - 400}{d^2}} $

$ \frac{60}{d} = \frac{40d}{d^2 - 400} $

Cross-multiply:

$ 60(d^2 - 400) = 40d^2 $

$ 60d^2 - 24000 = 40d^2 $

Solve for $d^2$:

$ 20d^2 = 24000 $

$ d^2 = 1200 $

Solve for $d$:

$ d = \sqrt{1200} = \sqrt{400 \times 3} $

$ d = 20\sqrt{3} $

The width of the road is $20\sqrt{3}$ m.