How do you solve the triangle given ∠B = 143°, ∠C = 20°, b = 37?

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How do you solve the triangle given ∠B = 143°, ∠C = 20°, b = 37?

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Answer 1 · 2016-08-04T04:56:39

$∠ A = 17^{0}$ $/_ A=17^0$ , $a \approx 18$ $a~~18$
$∠ B = 143^{0}$ $/_ B=143^0$ , $b = 37$ $b = 37$
$∠ C = 20^{0}$ $/_ C= 20^0$ , $c \approx 21$ $c~~21$

Explanation:

We kmow that $∠ B$ $/_B$ is $143^{o}$ $143^o$ and $C = 20^{o}$ $C=20^o$ . Since the angles within a triangle must add up to $180^{o}$ $180^o$ . $143 + 20 = 163$ $143+20=163$ . $180 - 163$ $180-163$ leaves us with $17^{o}$ $17^o$ . So now we know all the inner angles. We just need to find the lengths.

I'm going to use the law of sines, which says that $\frac{sin A}{a} = \frac{sin B}{b}$ $sinA/a=SinB/b$ . We can also write this as $\frac{a}{sin A} = \frac{b}{sin B}$ $a/SinA=b/SinB$ too since both of these formulas are synonymous. In order to use this formula we need three known variables and one unknown. We know the length of $b$ $b$ and all the angles, so we should be good.

Let's set this up:

$\frac{a}{{sin 17}^{o} =} \frac{37}{sin 143}$ $a/Sin17^o=37/Sin143$

$\frac{a}{.2923} = 61.481$ $a/.2923=61.481$

$a = 17.9752 \approx 18$ $a=17.9752~~18$

Great job! Let's do the next one!

$\frac{c}{{sin 20}^{o} =} \frac{37}{{sin 143}^{o}}$ $c/Sin20^o=37/Sin143^o$

$\frac{c}{.342} = 61.481$ $c/.342=61.481$

$c = 21.027 \approx 21$ $c=21.027~~21$

Okay, we're good! Nice work, we're done. We know all the lengths and all the angles.. Hopes this helps!

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How do you solve the triangle given ∠B = 143°, ∠C = 20°, b = 37?

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