Answer to Question #277258 in Trigonometry for Ola

Question

A vertical aerial AB 12.65m high stands on ground which is inclined 18°32 15to the horizontal. A stay connects the top of the aerial A to a point Con the ground 20.0 m uphill from Bthe foot of the aerial.

Determine:

the length of the stay .

the angle the stay makes with the ground

Answer 1

Given $AB=12.65m, BC=20.0m, \alpha=18\degree32'15''.$

$$\beta=90\degree-\alpha$$

Then

$$\angle ABC=\beta=90\degree-\alpha$$

Triangle ACB

$$AB=12.65, BC=20, ''$$

$$\angle ABC=90\degree-\alpha=90\degree-18\degree32'15''$$

a) Law of Cosines

$${AC}^2={AB}^2+{BC}^2-2(AB)(BC)\cos\angle ABC$$

$$AC=\sqrt{{12.65}^2+{20}^2-2(12.65)(20)\sin(18\degree32'15'')}$$

$$AC=19.98m$$

2) Law of Sines

$$\dfrac{\sin\angle ACB}{AB}=\dfrac{\sin \angle ABC}{AC}$$

$$\angle ACB=\sin^{-1}(\dfrac{AB\sin \angle ABC}{AC})$$

$$\angle ACB=\sin^{-1}(\dfrac{12.65\cos (18\degree32'15'')}{19.98})$$

$$\angle ACB=36.89\degree$$

$$\angle ACB=36\degree53'25''$$