How do you use law of sines to solve the triangle given C = 37°, a = 19, c = 8?

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How do you use law of sines to solve the triangle given C = 37°, a = 19, c = 8?

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Answer 1 · 2015-11-16T02:09:32

$A = {115.4}^{o}$ $A=115.4^o$
$B = {26.6}^{o}$ $B=26.6^o$
$b = 6.16$ $b=6.16$

Explanation:

$\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $(sinA)/a=(sinB)/b=(sinC)/c$

$\frac{sin A}{19} = \frac{sin C}{8}$ $(sinA)/19=(sinC)/8$
Multiply both sides by 19 and $sin A = 19 \frac{sin 37}{8}$ $sinA=19(sin37)/8$ which is $sin A = 1.429$ $sinA=1.429$
The next step is to take the ${sin}^{- 1}$ $sin^-1$ but you can not do an inverse of sin > 1 . We know that ${sin}^{- 1} (1) = 90^{o}$ $sin^-1(1)=90^o$ ${sin}^{- 1} (0.429) = 25.4$ $sin^-1(0.429)=25.4$ $90 + 25.4 = {115.4}^{o}$ $90+25.4=115.4^o$ That's your angle A.

Angle B is $180 - 115.4 - 37 = {27.6}^{o}$ $180-115.4-37=27.6^o$

$\frac{b}{sin 27.6} = \frac{8}{sin 37}$ $b/(sin27.6)=8/(sin37$ so $b = sin 27.6 (\frac{8}{sin 37})$ $b=sin27.6(8/(sin37))$
$b = 6.16$ $b=6.16$

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How do you use law of sines to solve the triangle given C = 37°, a = 19, c = 8?

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