Question

In the given problem solve the triangle using either the law of sines or law of cosines-

$b=5,c=12,A=60°$

Short answer

Required values of the triangle are

$a=10.44,B=13.8°,C=34.75°$

Step-by-step solution

Step 1. Given information

In the given triangle,

$b=5,c=12,A=60°$

We know Law of cosines,

$a^2=b^2+c^2-2bc\cos A$ -------$\left(1\right)$

Also, Law of sines,

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ -------$\left(2\right)$

Step 2. Calculation

For a, using equation $\left(1\right)$

$$\begin{gathered} a^2=b^2+c^2-2bc\cos A \\ \Rightarrow a^2=5^2+{12}^2-2\times 5\times 12.\cos 60° \\ \Rightarrow a^2=25+144-120\times 0.5 \\ \Rightarrow a^2=109 \\ \Rightarrow a=10.44 \end{gathered}$$

For B, using equation $\left(2\right)$

$$\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b} \\ \Rightarrow \frac{\sin 60°}{10.44}=\frac{\sin B}{5} \\ \Rightarrow \frac{0.5}{10.44}=\frac{\sin B}{5} \\ \Rightarrow \sin B=0.239 \\ \Rightarrow B={\sin }^{-1}\left(0.239\right) \\ \Rightarrow B=13.8° \end{gathered}$$

For C, using equation $\left(2\right)$

$$\begin{gathered} \frac{\sin A}{a}=\frac{\sin C}{c} \\ \Rightarrow \frac{\sin 60°}{10.44}=\frac{\sin C}{12} \\ \Rightarrow \sin C=\frac{0.5\times 12}{10.44} \\ \Rightarrow \sin C=0.57 \\ \Rightarrow C={\sin }^{-1}\left(0.57\right) \\ \Rightarrow C=34.75° \end{gathered}$$