Question

Three cell phone towers form a triangle. The distance between the

first tower and the second tower is 16 km. The distance between the
second tower and the third tower is 19 km. The distance between the
first tower and the third tower is 19 km. Calculate the angle formed
between the first cell phone tower and the third cell phone tower.
First, draw the top view diagram of this situation. Round your final
answers to the nearest degree, and state proper units.

Answer 1

`65^\circ`

Explanation:

Examine the attached diagram to follow the explanation

The locations of the 1st 2nd 3rd towers are labeled A, B, C

The distances are labeled a, b, c with
a = 16 km
b = 19 km
c = 19 km

The angle between AB and AC ie the angle formed between the first cell phone tower and the third cell phone tower is labeled
`\theta^\circ`

The law of cosines for a triangle states

`c^2 = a^2 + b^2 -2ab \cos(\theta)`

where c is the length of the side opposite
`\theta`

and a, b are the other two sides which form
`\angle \theta`

Substituting the given values for a, b, c and
`\theta` we get

`19^2 = 16^2 + 19^2 - 2(16)(19) \cos(\theta)\\\\19^2 \text{ cancels out since it appears on both sides of the equation}`

So we get

`0 = 16^2 - 2(16)(19) \cos(\theta)\\\\`

Add the term
`2(16)(19) \cos(\theta)` to both sides:

`2(16)(19) \cos(\theta) =16^2\\\\608 \cos(\theta) =256\\\\`

`\cos (\theta) = (256)/(608)\\\\\cos (\theta) = 0.42105\\\\\theta = cos^(-1) (0.42105)\\\\\theta = 65.09^\circ\\\\\theta = 65^\circ`
(when rounded to the nearest degree)