Given the sides, find the angles of the triangle.

Question

The problems says: "The sides of a triangle are $37$, $7$, $40$. Find all the angles, being given that $\cos 69^\circ 25' 48'' \approx 13/37$.

How I can use the fact that $\cos 69^\circ 25' 48'' \approx 13/37$ to find the angles of the triangle?

Answer 1

Huge hint:

The formula which most directly links the lengths of the sides of a triangle with the angles, when the sides are known and the angles are not, is the cosine formula $$a^2=b^2+c^2-2bc \cos A$$ where $A$ is the angle that is opposite the side of length $a$. You should try applying it. You may need some other knowledge about the cosines of "simple" angles, and about the cosines of obtuse angles. But you should be able to solve it from this.

Answer 2

Let's do the cosine rule:

$$40^2=37^2+7^2-2\cdot7\cdot37\cos x \Rightarrow \cos x=-\frac{13}{37} \Rightarrow x=180º-69º25'48''$$

You can proceed from here.