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#1. Use the law of sines to solve ASA or AAS triangles (Knewton 10.1) In AABC, we are told that b = 13, 2B = 54°, and 20 – 43
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\small \\$Given that in $\triangle ABC\\ \angle B=54\degree,\angle C=43\degree,b=13\\ $We know that the sum of angles of a triangle is $180\degree\\ So,\angle A=180\degree-(\angle B+\angle C)=180\degree-(54\degree+43\degree)=83\degree\\ $We know from sine rule$\\ \frac{a}{sin\angle A}=\frac{b}{sin\angle B}=\frac{c}{sin\angle C}\\\\ \Rightarrow \frac{a}{sin(83\degree)}=\frac{13}{sin(54\degree)}=\frac{c}{sin(43\degree)}\\\\ So, a=\frac{13sin(83\degree)}{sin(54\degree)}=15.94910868\approx 15.95\\ c=\frac{13sin(43\degree)}{sin(54\degree)}=10.95895233\approx10.96

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