When given an angle, the side opposite the angle, and another side, you have to determine how many triangles are possible or if it is not possible.
Begin with when a triangle cannot exist:
1.) measure of non-inclusive angle is less than 90° and side opposite the angle < (other side)×sin(angle measure)
2.) measure of non-inclusive angle is ≥ 90° and side opposite the angle < other side
Next determine if two triangles exist:
Only if the measure of angle is less than 90° and
(other side)×sin(non-inclusive angle measure) < opposite side < other side
Otherwise 1 triangle exist...
HERE IS WHAT YOUR PROBLEM HAS:
Non-inclusive angle measure = 22° which is < 90°
Opposite side = 13
Other side = 18.1
(other side)×sin(non-inclusive angle measure) =(18.1)×sin(22°) ≈ 6.78
So how many triangles?
6.78 < 13 < 18.1 so 2 triangles exist
Now let's find them... find angle B with law of sines
[tex] \frac{sin22^o}{13} = \frac{sinB}{18.1} [/tex]
Put the following in your calculator: [tex]sin^{-1}( \frac{18.1sin(22^o)}{13})[/tex]
This gives you the first angle B value and if you subtract it from 180 you get the other angle B value
To find the angle C for the first triangle 180 - (sum of angle A and first angle B)
Then use the law of sines to find side c for first triangle.
To find the angle C for the 2nd triangle 180 - (sum of angle A and 2nd angle B)
Then use the law of sines to find side c for 2nd triangle.
Sorry, I didn't do the calculations because my calculator is dead.
