Question

Using the law of sines, determine whether the given information results in one triangle, two triangle or no triangle at all. Solve any triangle that results. a=8angle B = 57 degreeAngle A =49 degree

Answer 1

step 1

Find out the measure of angle C

Remember that

In any triangle, the sum of the interior angles must be equal to 180 degrees

so

A+B+C=180 degrees

substitute given values

49+57+C=180

C=180-106

C=74 degrees

step 2

Find out the length side b

Applying the law of sines

$\frac{a}{sinA}=\frac{b}{sinB}$

substitute

$\frac{8}{s\imaginaryI n49^o}=\frac{b}{s\imaginaryI n57^o}$

Solve for b

$\begin{gathered} b=\frac{8*s\mathrm{i}n57^o}{s\imaginaryI n49^o} \\ \\ b=8.89\text{ ---> rounded to two decimal places} \end{gathered}$

step 3

Find out the length side c

Applying the law of sines

$\frac{a}{s\imaginaryI nA}=\frac{c}{s\imaginaryI nC}$

substitute

$\frac{8}{s\imaginaryI n49^o}=\frac{c}{s\imaginaryI n74^o}$

solve for c

$\begin{gathered} c=\frac{8*s\mathrm{i}n74^o}{s\imaginaryI n49^o} \\ \\ c=10.19\text{ ----> rounded to two decimal places} \end{gathered}$

therefore

N of triangles is only one

b=8.89

c=10.19

C=74 degrees