Solve the triangle.

Question

Consider a triangle ABC, . Suppose a=67, b=73, and c= 12

. (The figure is not drawn to scale.) Solve the triangle.

Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.

If there is more than one solution, use the button labeled "or".

Answer 1

Let a = 67, b=12 and c=73 opposite angles A, B and C respectively. Use the law of cosines to find angle C:

c² = a² + b² -2abcos(C)

2abcos(C) = a + b² – c²

cos(C) = (a² + b² – c²)/2ab

cos(C) = (672 + 122 – 732)/(2*67*12)

cos(C) = -.04328

C = 115.6376º

Then using the law of sines you can get the other two angles B and A.

sin(C)/c = sin(B)/b

sin(B) = b*sin(C)/c

sin(B) = (12*sin(115.6376º))/73

sin(B) = 0.1482

B = 8.5219º

sin(C)/c = sin(A)/a

sin(A) = a*sin(C)/c

sin(A) = (67*sin(115.6376º))/73

sin(A) = 0.8274

A = 55.8305º

Sum of angles is:

A + B + C = 55.8305º + 8.5219º + 115.6376º = 179.99º ~ 180º

Rounding answers to nearest tenth digits gives :

A = 55.8º, B = 8.5º and C = 115.6º. (Note rounding C to 115.7º would give 180º as the sum.)