Find the angles and the sides of a triangle with a=7, b=5, and B=40 deg.

We have SSA (the ambiguous case) here..

Using the Law of Sines, we have

sinA / a = Sin B / b

sinA / 7 = sin40 / 5

sinA = (7/5)sin40

sin^-1((7/5)sin 40) =A = about 64.15 degrees

So, angle C is 180 - 40 - 64.15 = 75.85 degrees

And side C is given by

a / sin A = c / sin C

5 / sin 40 = c / sin 75.85 degrees

c = 5 sin75.85 / sin 40  = 7.54

But....angle A could also be a second quadrant angle = 180 - 64.15 = 115.85

So the other possible angles are 40, 115.85 and 24.15

And side c in this triangle is given by

5 / sin 40 = c / sin 24.15

c = 5sin24.15/sin40 = 3.18

So the two triangle are (sides, angles)

a = 7, b = 5,  c = 7.54    A = 64.15, B = 40, C = 75.85   and

a = 7, b = 5, c = 3.18    A= 115.85, B = 40, C = 24.15

Draw the triangle

use the sine rule to find angle A

use angle sum of  a triangle to find angle c

use sine or cosine rule to find side c