How do you solve the triangle given angle A=32degrees, C=44degrees, c=20?

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How do you solve the triangle given angle A=32degrees, C=44degrees, c=20?

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Answer 1 · 2015-11-16T20:38:29

$B = 104^{o}$ $B=104^o$
$a = 15.26$ $a=15.26$
$b = 27.94$ $b=27.94$

Explanation:

First solve for angle B by subtracting your known angles (32 and 44) from 180. $180^{o} - 32^{o} - 44^{o} = 104^{o}$ $180^o-32^o-44^o=104^o$

Then use law of sines to solve for a and b. $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $(sinA)/a=(sinB)/b=(sinC)/c$

$\frac{b}{sin 104} = \frac{20}{sin 44}$ $b/sin104=20/(sin44)$ multiply both sides by sin104
$b = sin 104 (\frac{20}{sin 44}) = 27.94$ $b=sin104(20/(sin44))=27.94$

$\frac{a}{sin 32} = \frac{20}{sin 44}$ $a/sin32=20/(sin44)$
$a = sin 32 (\frac{20}{sin 44}) = 15.26$ $a=sin32(20/(sin44))=15.26$

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How do you solve the triangle given angle A=32degrees, C=44degrees, c=20?

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