Question

Solve the triangles with the given parts: a=103, c=159, m∠C=104º

Answer 1

Sides:

• 
  `a= 103`.
• 
  `b \approx 99`.
• 
  `c - 159`.

Angles:

• 
  `\angle A \approx 39^\circ`.
• 
  `\angle B \approx 37^\circ`.
• 
  `\angle C = 104^\circ`.

Explanation:

Angle A

Apply the law of sines to find the sine of
`\angle A`:

`\displaystyle (sin(A))/(sin(C)) = (a)/(c)`.

`\displaystyle\sin A = (a)/(c) \cdot sin(C) = (103)/(159) * \left(\sin{104^(\circ)}\right) \approx 0.628556`.

Therefore:

`\angle A = \displaystyle\arcsin (\sin A) \approx \arcsin(0.628556) \approx 38.9^\circ`.

Angle B

The three internal angles of a triangle should add up to
`180^\circ`. In other words:

`\angle A + \angle B + \angle C = 180^\circ`.

The measures of both
`\angle A` and
`\angle C` are now available. Therefore:

`\angle B = 180^\circ - \angle A - \angle C \approx 37.1^\circ`.

Side b

Apply the law of sines (again) to find the length of side
`b`:

`\displaystyle(b)/(c) = (\sin \angle B)/(\sin \angle C)`.

`\displaystyle b = c \cdot \left((\sin \angle B)/(\sin \angle C)\right) \approx 159* (\sin \left(37.1^\circ\right))/(\sin\left(104^\circ\right)) \approx 98.8`.