Solve the following triangle using either the Law of Sines or the Law of Cosines. A= 13 degrees, a=9, b=12. B1= C1= C2= b2= c1= c2=
Added by Jose Ramon H.
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We have $\frac{a}{\sin A} = \frac{b}{\sin B}$, so we can solve for $\sin B$: $\sin B = \frac{b \sin A}{a} = \frac{12 \sin 13^\circ}{9}$ Now, we can find angle B: $B = \sin^{-1}\left(\frac{12 \sin 13^\circ}{9}\right) \approx 22.07^\circ$ Show more…
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