How do you solve the triangle ABC, given a=15, A=94, b=12?

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How do you solve the triangle ABC, given a=15, A=94, b=12?

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Answer 1 · 2017-01-04T04:13:18

Angle A is obtuse, therefore, the ambiguous case does not apply and there is only one possible triangle. Use the Law of Sines to solve.

Explanation:

Use the Law of Sines :

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

$B = {sin}^{- 1} (\frac{b}{a} sin (A))$ $B = sin^-1(b/asin(A))$

$B = {sin}^{- 1} (\frac{12}{15} sin (94^{\circ}))$ $B = sin^-1(12/15sin(94^@))$

$B \approx {52.9}^{\circ}$ $B ~~ 52.9^@$

$C \approx 180^{\circ} - 94^{\circ} - {52.9}^{\circ}$ $C ~~ 180^@ - 94^@ - 52.9^@$

$C \approx {33.1}^{\circ}$ $C ~~ 33.1^@$

$c = a \frac{sin (C)}{sin (A)}$ $c = asin(C)/sin(A)$

$c \approx 8.2$ $c ~~ 8.2$

The solution for the triangle is:

$a = 15, b = 12, c \approx 8.2, A = 94^{\circ}, B \approx {52.9}^{\circ}, and C \approx {33.1}^{\circ}$ $a = 15, b = 12, c ~~ 8.2, A = 94^@, B ~~ 52.9^@, and C ~~ 33.1^@$

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How do you solve the triangle ABC, given a=15, A=94, b=12?

1 Answer

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