Solve each triangle described below.Round measures of sides and angles to the nearest tenth.
1. B=71° , c=8 , b=16
2. A=40° , B=14° , a=52
+12530 Solve each triangle described below.Round measures of sides and angles to the nearest tenth.
+130087 1. B =71°, c = 8, b = 16
We have a SSA situation.....so.....we will have either 1 triangle, 2 triangles or 0 triangles
Using the Law of Sines
sin 71 / 16 = sin C / 8
sin C = .5sin71
arcsin (.5sin 71) = C ≈ 28.2° or 180 - 28.2 = 151.8°
Note that the second angle value isn't possible since 71 + 151.8 > 180°
So....only one triangle exists
A = 180 - 71 - 28.2 = 80.8°
And
a/ sin (80.8) = 16/sin 71
a = 16sin (80.8)/sin 71 ≈ 16.7
A = 80.8° a = 16.7
B = 71° b = 16
C = 28.2° c = 8
