Question

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

Answer 1

Answer:

Sides:

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

Angles:

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

Step-by-step explanation:

Angle A

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

$\displaystyle \frac{\sin{A}}{\sin{C}} = \frac{a}{c}$.

$\displaystyle\sin A = \frac{a}{c} \cdot \sin{C} = \frac{103}{159} \times \left(\sin{104^{\circ}}\right) \approx 0.628556$.

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 $b$:

$\displaystyle\frac{b}{c} = \frac{\sin \angle B}{\sin \angle C}$.

$\displaystyle b = c \cdot \left(\frac{\sin \angle B}{\sin \angle C}\right) \approx 159\times \frac{\sin \left(37.1^\circ\right)}{\sin\left(104^\circ\right)} \approx 98.8$.