Calculating $AB$

Question

Here is my diagram

I want to calculate $AB$ and don't have any idea about where to start. Might I get any help for this question?

Regards!

Answer 1

In this right triangle, by definition$$sin37^o=\frac{AB}{AC}$$So$$AB=ACsin 37^o$$Look up $sin 37^o$, and then take the product $ACsin 37^o$. When $BC$ rather than $AC$ is given along with $\angle C$, you use $tanC$ to find $AB$, which is a good way to compute the height of things you'd rather not--or can't--measure by climbing them.

Answer 2

Hint : First use the fact that the angles of a triangle always add up to 180º (you know two angles already).

Once you know all the angles, use the Law of Sines:

$$\dfrac{a}{sin \ A}=\dfrac{b}{sin \ B} = \dfrac{c}{sin \ C}$$