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.
Similar Questions
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Statement p :
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Statement q :
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Important Questions from Trigonometry
Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
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