How do you solve the triangle ABC, given b=40, B=45, c=15?

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How do you solve the triangle ABC, given b=40, B=45, c=15?

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Answer 1 · 2016-11-03T02:54:40

$∠ A = 120^{\circ}$ $angleA=120^circ$
$∠ C \approx 15^{\circ}$ $angleC~~15^circ$
$a \approx 49$ $a~~49$

Explanation:

This is an SSA triangle so you can use the Law of Sine to solve this
Law of Sine = $\frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}$ $sinA/a=sinB/b=sinC/c$
in this case we use $\frac{sin B}{c} = \frac{sin C}{c}$ $sinB/c=sinC/c$
Given: $b = 40, B = 45^{\circ}, c = 15$ $b=40, B=45^circ, c=15$

first let's find $∠ C$ $angleC$
$\frac{{sin 45}^{\circ}}{40} = \frac{sin C}{15}$ $sin45^circ/40=sinC/15$

$15 {sin 45}^{\circ} = 40 sin C$ $15sin45^circ=40sinC$

$∠ C = {sin}^{- 1} (\frac{15 {sin 45}^{\circ}}{40}) \approx 15^{\circ}$ $angleC=sin^-1((15sin45^circ)/40)~~ 15^circ$
$∠ C \approx 15^{\circ}$ $angleC~~15^circ$

now we have $∠ B$ $angleB$ and $∠ C$ $angleC$ so we can subtract both from
$180^{\circ}$ $180^circ$ to find the third angle
$∠ A = 180^{\circ} - (15^{\circ} + 45^{\circ}) = 120^{\circ}$ $angleA=180^circ-(15^circ+45^circ)=120^circ$
$∠ A = 120^{\circ}$ $angleA=120^circ$

to find $a$ $a$ we use Law of Sine again
$\frac{{sin 45}^{\circ}}{40} = \frac{{sin 120}^{\circ}}{a}$ $sin45^circ/40=sin120^circ/a$
$a {sin 45}^{\circ} = 40 {sin 120}^{\circ}$ $asin45^circ=40sin120^circ$
$a = \frac{40 {sin 120}^{\circ}}{{sin 45}^{\circ}} \approx 49$ $a=(40sin120^circ)/(sin45^circ)~~49$
$a \approx 49$ $a~~49$

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How do you solve the triangle ABC, given b=40, B=45, c=15?

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