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The horizontal distance between two stations P and Q is 100 m. The vertical angles from P and Q to the top of a vertical tower at T are 3° and 5° above horizontal, respectively. The vertical angles from P and Q to the base of the tower are 0.1° and 0.5° below horizontal, respectively. Stations P, and the tower are in the same vertical plane with P and Q being on the same side of T. Neglecting earth’s curvature and atmospheric refraction, the height (in m) of the tower is
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- 6.972
- 12.387
- 12.540
- 128.745
- None of the above
- 6.972
Correct Option: E
None of the options is correct.
Let QC = x
| In ∆AQC, | = tan 5° .....(i) | |
| x |
| In ∆APC, | = tan 3° .....(ii) | |
| x + 100 |
| In ∆BQC, | = tan 0.5° .....(iii) | |
| x |
| In ∆BPC, | = tan 1° .....(iv) | |
| x + 100 |
Height of tower = h1 + h2
Solving (i) and (ii) we get, x = 149.39 m
Substituting in (i) We get, h1 = 13.07 m
Substituting in (iii) we get, h2 = 1.303 m
∴ height of h = h1 + h2 = 14.37 m
