Three towns (A,B, and C) are located so that B is 25 km from A and C is 34 km from A. If <ABC is 110 degrees, how do you calculate the distance from B to C?

1 Answer
Aldrich
Sep 7, 2015

x=(34sin (70 - sin^-1((25sin110)/34)))/sin110

Explanation:

First, let's draw the triangle in question using all the given.

![Not to http://scale.](https://useruploads.socratic.org/Ma6BTNgXR8Sfh2T1PSBw_triangle.png)
In this diagram, a and c are angles, and x is a side length.

By the Law of Sines, we know that:

sin a/x = sin110/34 = sinc/25

We can immediately solve for c.

sin110/34 = sinc/25

(25sin110)/34=sinc

c=sin^-1((25sin110)/34)

The sum of angles in a triangle is 180, so we can solve for a:

180 = a + 110 + sin^-1((25sin110)/34)

a = 70 - sin^-1((25sin110)/34)

We now just need to solve the following equation for x:

sin (70 - sin^-1((25sin110)/34))/x = sin110/34

(34sin (70 - sin^-1((25sin110)/34)))/sin110 = x

None of the angles or values used are "standard" values on the unit circle, so this is the final answer.

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