There are two triangles satisfying \(\angle BCA = 60^{\circ}, AB = 9\) and \(AC = 10\) shown below:
The possible values of BC are a1 and a2, as shown in the picture, with a1
a1, a2
in that order?
We have an example of the ambiguous case = SSA
Using the Law of Sines
sin B / 10 = sin C / 9
sin B /10 = [sqrt (3) /2] / 9
sin B = 10 sqrt (3) / 18 = (5/9)sqrt 3 =
arcsin [ (5/9)sqrt 3 ] = B ≈ 74.2° or B ≈ 105.8°
So
So , in one case, C = 180 - 60 - 74.2 ≈ 45.8°
So...let's figure a2 first
a2 / sin (45.8) = 9 /sin C
a2 / sin (45.8) = 9 / [ sqrt (3) / 2 ]
a2 =sin (45.8) * 18 / sqrt 3 ≈ 7.45
And in the second case
C = 180 -60 - 105.8 ≈ 14.2°
So....by the Law of Sines again we can find a1 as
a1 / sin (14.2) = 18/sqrt 3
a1= sin (14.2) * 18 /sqrt 3 ≈ 2.55
{a1 , a2} = {2.55 , 7.45}