Question

Using the law of sines, determine whether the given information results in one triangle, two triangles or no triangle at all. Solve any triangle (s) that results. See picture for details

Answer 1

$\begin{gathered} b=8.89 \\ c=10.18 \\ \angle C=74\text{ \degree} \end{gathered}$

Explanation

the law of sines states that

$\frac{\sin\text{ A}}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

hence

Step 1

a)let

$\begin{gathered} a=8 \\ \angle B=57\text{ \degree} \\ \angle A=\angle49\text{ \degree} \end{gathered}$

b) replace to find B

$\begin{gathered} \frac{\sin\text{ A}}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ \frac{\sin49}{8}=\frac{\sin(57)}{b} \\ b*sin49=8sin57 \\ b=\frac{8sin\text{ 57}}{sin\text{ 49}} \\ b=8.89 \end{gathered}$

c) we can find the angle C usign the fact that the sum of the internal angles in a triangle equals 180,so

$\begin{gathered} \angle A+\angle B+\angle C=180 \\ replace\text{ and solve for }\angle \\ 49+57+\angle C=180 \\ \angle C=180-49-57 \\ \angle C=74 \end{gathered}$

c) finally, side c

$\begin{gathered} \frac{\sin\text{ A}}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ \frac{\sin(\text{A})}{a}=\frac{\sin(C)}{c} \\ c*sin(A)=asin(C) \\ c=\frac{asin\text{ \lparen C\rparen}}{sin\text{ \lparen A\rparen}} \\ replace \\ c=\frac{8sin(74)}{sin(49)} \\ c=10.18 \end{gathered}$

so, the answer is ( one triangle)

$\begin{gathered} b=8.89 \\ c=10.18 \\ \angle C=74\text{ \degree} \end{gathered}$

I hope this helps you