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In △ABC, m∠A=55°, c=11, and m∠B=19°. Find the perimeter of the triangle.law of sines 4A. 39B. 24C. 32D. 48

Sagot :

Solution: 24.096

Analysis:

We have a triangle and know two angles and one side of the triangle. According to the angle rule, a triangle's total intern angles equals 180 degrees. We have:

mm∠B=19°

∠A+∠B+∠C=180°

55°+19°+∠C=180°

∠C=180°-55°-19°

∠C=106°

[tex]\frac{a}{sin(A)}=\frac{b}{sin(B)}=\frac{c}{sin(C)}[/tex][tex]\begin{gathered} \frac{a}{sin(55)}=\frac{11}{sin(106)} \\ \\ a=\frac{11\ast sin(55)}{sin(106)}=\frac{11\ast0.82}{0.96}=9.37 \end{gathered}[/tex]

[tex]\begin{gathered} \frac{b}{sin(19)}=\frac{11}{sin(106)} \\ \\ b=\frac{11\ast sin(19)}{sin(106)}=\frac{11\ast0.325}{0.96}=3.726 \end{gathered}[/tex]

Perimeter=a+b+c

Perimeter=9.37+3.726+11

Perimeter=24.096