# How do you find the missing sides of a triangle given two angles and a side given Sides are x, y, and 10, two of the angles are 40 and 90?

May 2, 2017

If $10$ = hypotenuse: $x \approx 6.43 , y \approx 7.66$
If $10$ is across from ${50}^{\circ} , x \approx 8.39 , y \approx 13.05$
If $10$ is across from the ${40}^{\circ} , x \approx 11.92 , y \approx 15.56$ bolded text

#### Explanation:

Given: A triangle has two angles: ${40}^{\circ} , {90}^{\circ}$, and side lengths x, y, & 10.

Since we have a ${90}^{\circ}$ angle you know that you have a right triangle, which means you can use trigonometric functions.

There are three possible cases:

The longest side is always across from the largest angle. This means that $10$ is the hypotenuse if 10 is the longest side.

1. Assume $10$ is the hypotenuse:

Since a triangle's angles sum to ${180}^{\circ}$, the third angle is ${50}^{\circ}$.

cos 40^@ = y/10; " " y = 10 cos 40^@ ~~ 7.66

sin 40^@ = x/10; " " x = 10 sin 40^@ ~~ 6.43

1. Assume $10$ is across from the ${50}^{\circ}$ angle:

tan 40^@ = x/10; " " x = 10 tan 40^@ ~~ 8.39

cos 40^@ = 10/y; " " y = 10/(cos 40^@) ~~ 13.05

1. Assume $10$ is across from the ${40}^{\circ}$ angle:

tan 40^@ = 10/x; " "x = 10/(tan 40^@) ~~ 11.92

sin 40^@ = 10/y; " " y = 10/(sin 40^@) ~~ 15.56#