Question

In triangle ABC, a=9, c=5, and angle B = 120 degrees. What is the measure of angle A to the nearest degree?

Answer 1

About 39 degree.

Explanation:

First, calculate the edge `b` with the law of cosine.

`b^2=a^2+c^2-2ac cosB`
`=9^2+5^2-2*5*9*(-1/2)=151`
`b=sqrt(151)`

Then, calculate the angle `A` with the law of sine.
`a/(sinA) = b/(sinB)`
`asinB=bsinA`
`sinA=(asinB)/b=(9*sqrt(3)/2)/sqrt(151)`
`=(9sqrt(453))/302`
`≒0.6343`

Note that `A` must be smaller than `180-120=60` degree.
Thus, angle `A` is
`A=arcsin(0.6343)≒39.37 deg`.