# How do you find the angles of a triangle given sides 9, 6, and 14?

Jul 8, 2017

The angles are 137°, 26° and 17°

#### Explanation:

Firstly, Is it a right-angled triangle?
Check using Pythagoras's Theorem.
${6}^{2} + {9}^{2} = 117 \text{ and } {14}^{2} = 196$
No 90° angle, but the biggest angle will be obtuse because the square of the longest side is bigger than the sum of the squares of the other 2 sides.

Use the Cosine rule to find the biggest angle first - that will tell you whether the angle is acute or obtuse.

$C o s \theta = \frac{{6}^{2} + {9}^{2} - {14}^{2}}{2 \times 6 \times 9} = - 0.73148$
theta = 137°

Now use the Sine rule:

$\frac{S \in \beta}{9} = \frac{\sin 137}{14}$

$\sin \beta = \frac{9 \sin 137}{9} = 0.438339$
beta = 26°

Use the sum of the angles to find the third angle:

gamma = 180°-137°-26° =17°