Question

Use the Law of Sines to solve the triangle. Round your answers to two decimal places.A = 8° 40', B = 13° 15', b = 4.8

Answer 1

Given

$A=8°40^{\prime},B=13°15^{\prime},b=4.8$

To find the value of a, c, C.

Explanation:

It is given that,

$A=8°40^{\prime},B=13°15^{\prime},b=4.8$

Since,

$A=8°40^{\prime},B=13°15^{\prime}$

Then,

$\begin{gathered} A+B+C=180 \\ 8\degree40^{\prime}+13\degree15^{\prime}+C=180\degree \\ C=180\degree-21\degree55^{\prime} \\ C=(179-21)\degree(60^-55^)^{\prime} \\ C=158\degree5^{\prime} \end{gathered}$

Therefore, by using Sine law,

$\begin{gathered} \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ \frac{\sin8\degree40^{\prime}}{a}=\frac{\sin13\degree15^{\prime}}{4.8}=\frac{\sin158\degree5^{\prime}}{c} \\ \Rightarrow\frac{\sin8\degree40^{\prime}}{a}=\frac{\sin13\degree15^{\prime}}{4.8} \\ \Rightarrow\frac{\sin13\degree15^{\prime}}{4.8}=\frac{\sin158\degree5^{\prime}}{c} \end{gathered}$

Therefore,

$\begin{gathered} \begin{equation*} \frac{\sin8\degree40^{\prime}}{a}=\frac{\sin13\degree15^{\prime}}{4.8} \end{equation*} \\ \frac{0.14608}{a}=\frac{0.227501}{4.8} \\ a=\frac{0.14608}{0.047396} \\ a=3.08211 \\ a=3.1 \end{gathered}$

Also,

$\begin{gathered} \begin{equation*} \frac{\sin13\degree15^{\prime}}{4.8}=\frac{\sin158\degree5^{\prime}}{c} \end{equation*} \\ 0.047396=\frac{0.366501}{c} \\ c=\frac{0.366501}{0.047396} \\ c=7.73274 \\ c=7.7 \end{gathered}$

Hence, the answer is

$C=158\degree5^{\prime},a=3.1,c=7.7$