Solve the triangle given two side lengths and one angle measure.

Step-by-step explanation:

The Law of Sines is helpful when you know one side and its opposite angle.

  a/sin(A) = b/sin(B) = c/sin(C)

Rearranging gives you ...

  B = arcsin(b/a·sin(A)) = arcsin(105/109·sin(70°)) ≈ 64.85138°

  C = 180° -B -A = 45.14862°

  c = a·sin(C)/sin(A) ≈ 82.23360

Comment on the solution method

You can use the Law of Cosines if you like. The formulation would be ...

  a² = b² + c² -2bc·cos(A) . . . . where a, b, and A are known

This gives a quadratic in c, the positive solution being the answer you're looking for. Then, either the law of sines or the law of cosines can be used to find one of the other two angles.

  c = 105·cos[70°] + √[856 + 11025·cos[70°]²]

  c ≈ 82.2336

multiple all that

step-by-step explanation:

27×17