Question

The prompt's to solve the triangle. I've gotten the side lengths down, but I'm not too sure about the angle measures. Can anyone help?

Answer 1

Given:

$\begin{gathered} b=7 \\ c=3 \\ \angle A=41^{\circ} \end{gathered}$

To find: The missing sides and angles

Explantion:

The formula of cosine,

$a=\sqrt[]{b^2+c^2-2bc\cos A}$

Substituting the given values we get,

$\begin{gathered} a=\sqrt[]{3^2+7^2-2(3)(7)\cos 41^{\circ}} \\ =\sqrt[]{9+49-42\cos41^{\circ}} \\ =\sqrt[]{26.3021} \\ =5.12857 \\ \approx5.1 \end{gathered}$

Thus, the missing side length is a = 5.1.

Using the sine formula,

$\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ \frac{\sin41^{\circ}}{5.1}=\frac{\sin B}{3}=\frac{\sin C}{7} \end{gathered}$

Solving first two terms we get,

$\begin{gathered} \frac{\sin41^{\circ}}{5.1}=\frac{\sin B}{3} \\ \sin B=\frac{3\sin41^{\circ}}{5.1} \\ =\sin ^{-1}\mleft(0.39\mright) \\ =22.70 \\ B\approx23^0 \end{gathered}$

Thus, the measure of angle B is 23 degrees.

Using the angle sum property,

$\begin{gathered} A+B+C=180^{\circ} \\ 41+23+C=180 \\ C=180-64 \\ C=116^{\circ} \end{gathered}$

Thus, the measure of angle C is 116 degrees.

Final answer: The missing angles and sides are,

$\begin{gathered} \angle B=23^{\circ} \\ \angle C=116^{\circ} \\ BC=a=5.1 \end{gathered}$