How do you find $∠ B$ $angleB$ if in triangle ABC, $a = 15$ $a = 15$ , $b = 20$ $b = 20$ , and $∠ A = 30^{\circ}$ $angle A=30^@$ ?

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How do you find $∠ B$ $angleB$ if in triangle ABC, $a = 15$ $a = 15$ , $b = 20$ $b = 20$ , and $∠ A = 30^{\circ}$ $angle A=30^@$ ?

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Answer 1 · 2014-12-17T21:17:32

The answer is: $∠ B = {sin}^{- 1} (\frac{20 \cdot sin (30^{\circ})}{15})$ $/_B = sin^-1((20*sin(30^∘))/15)$

You have to use the law of sines, just as you thought. This law is:
$\frac{a}{sin ∠ A} = \frac{b}{sin ∠ B} = \frac{c}{sin ∠ C}$ $a/(sin/_A) = b/(sin/_B) = c/(sin/_C)$
We only need the first equation here, let's fill in the numbers.
$\frac{15}{sin (30^{\circ})} = \frac{20}{sin ∠ B}$ $15/sin(30^∘) =20/(sin/_B)$
We use cross-multiplication, then we find:
$sin ∠ B = \frac{20 \cdot sin (30^{\circ})}{15}$ $sin/_B = (20*sin(30^∘))/15$
then by taking the inverse sine, we find
$∠ B = {sin}^{- 1} (\frac{20 \cdot sin (30^{\circ})}{15})$ $/_B = sin^-1((20*sin(30^∘))/15)$

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How do you find $∠ B$ $angleB$ if in triangle ABC, $a = 15$ $a = 15$ , $b = 20$ $b = 20$ , and $∠ A = 30^{\circ}$ $angle A=30^@$ ?

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How do you find angleB∠BangleB if in triangle ABC, a = 15a=15a = 15, b = 20b=20b = 20 , and angle A=30^@∠A=30∘angle A=30^@?

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How do you find $∠ B$ $angleB$ if in triangle ABC, $a = 15$ $a = 15$ , $b = 20$ $b = 20$ , and $∠ A = 30^{\circ}$ $angle A=30^@$ ?