Question

Consider a triangle ABC like the one below. Suppose that B = 46°, C = 72°, and a = 74. (The figure is not drawn to scale.) Solve the triangleRound your answers to the nearest tenth.If there is more than one solution, use the button labeled "or".

Answer 1

A triangle has 3 sides and 3 angles

Angles

B = 46 degrees (given)

C = 72 degrees

To find angle A

A + 46 + 72 = 180 (sum of angles in a triangle)

A + 118 = 180

A = 180 - 118

A=62

A = 62.0 degrees (to the nearest tenth)

Sides

a = 74 (given)

To find side b and c

We use sine rule

$\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \frac{\sin62}{74}=\frac{\sin46}{b} \\ \text{Cross multiplying, we have} \\ b\text{ x sin62 = 74 x sin 46} \\ b=\frac{74\text{ x 0.7193}}{0.883} \\ b=60.28 \end{gathered}$

Therefore b = 60.3 (to the nearest tenth)

$\begin{gathered} \frac{\sin C}{c}=\frac{\sin A}{a} \\ \frac{\sin 72}{c}=\frac{\sin 62}{74} \\ \text{Cross multiplying we have,} \\ c\text{ x sin62 = 74 x sin72} \\ c=\frac{74\text{ x 0.9511}}{0.883} \\ c=79.71 \end{gathered}$

Therefore c = 79.7 (to the nearest tenth)