triangle ABC has b = 10, c = 15, and m angle C = 80 degrees

Question

i am trying to solve the following triangles.

Answer 1

Given triangle sides b = 10, c = 15, C = 80

To solve triangle find the side a and angles B,A.

From the Sin rule a/SinA = b/SinB = c/SinC

b/SinB = c/SinC

10/SinB = 15/Sin(80)

10/SinB = 15/0.98

10/SinB = 15.306

SinB = 10/15.306

SinB = 0.653

B = Sin^-1(0.653)

B = 40.76

B = 41

A = 180-(B+C)

A = 180-(41+80)

A = 59

From the Cosine law a^2 = b^2+c^2-2bcCosA

a^2 = (10)^2+(15)^2-2(10)(15)Cos(59)

a^2 = 100+225-300(0.51)

a^2 = 325-153

a^2 = 172

a = Sqrt(172)

a = 13.11

Remaining side of the triangle a is 13.11 and angle A = 59, B = 41.