Question

A tower 150 m high is situated at the top of a hill. At a point 650 m down the hill, the angle
between the surface of the hill and the line of sight to the top of the tower is 12 deg 30 min.
Find the inclination of the hill to a horizontal plane.

Answer 1

Let

A = top of the tower,

B = base of the tower and

C = point of observation

=> ∠ ACB = 12.5°

Let θ = angle of inclination of the hill to the horizontal plane.

=> ∠ CAB = 90° - (12.5° + θ)

Applying sine rule to the triangle ABC,

sin [90° - (12.5° + θ)] / 650 = sinθ / 150

=> cos (12.5° + θ) / sinθ = 650/150

=> cos(12.5°) cotθ - sin(12.5°) = 13/3

=> (0.9763) cotθ = 4.3333 + 0.2164

=> cotθ = 4.6602

=> θ = 12.11°
- Kesha

wait, what's the answer choices???

Answer 2

do the choices happen to be:
A. 5°54'

B. 7°10'

C. 6°12'

D. 7°50
?

Answer 3

1/3

because if so, it would be D)
Reasoning:

Applying sine rule to the triangle ABC,

sin [90° - (12.5° + θ)] / 650 = sin(12.5°) / 150

=> cos (12.5° + θ) / sin(12.5°) = 650/150

=> cot(12.5°) cosθ - sinθ = 13/3

=> 4.5107 cosθ - sinθ = 4.3333

=> 4.5107 (1 - tan^2 (θ/2)) - 2tan(θ/2) = 4.3333 (1 + tan^2 (θ/2))

=> 8.841 tan^2 (θ/2) + 2tan(θ/2) - 0.1774 = 0

=> tan(θ/2) = 1/(17.682) [- 2 + √(4 + 6.2736)]

=> tan(θ/2) = 0.06816

=> θ/2 = 3.9°

=> θ = 7.8°.

Answer: D).

Answer 4

From your:
cos (12.5° + θ) / sin(12.5°) = 650/150

why not just:
cos(12.5+θ) = 650sin12.5/150 = .9379...
12.5+θ = 20.2973...
θ = 7.797
or their answer of 7° 50'