How do you solve the triangle using law of sines given a=8, b=9, c=10?

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How do you solve the triangle using law of sines given a=8, b=9, c=10?

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Answer 1 · 2015-06-04T17:20:26

Given three sides of a triangle, angles can be found using cosine rule ans not the Rule of sines. Accordingly,cosA= $\frac{b^{2} + c^{2} - a^{2}}{2 b c}$ $(b^2+c^2-a^2)/(2bc)$ = $\frac{81 + 100 - 64}{180} = \frac{17}{180} . A = a r c cos (\frac{117}{180})$ $(81+100-64)/180= 17/180. A= arc cos(117/180)$ = 49.45degrees. Like wise cosB= $\frac{83}{160}$ $83/160$ . B= arc cos(83/160)=58.75 degrees. Then finally C= 180-49.45-58.75= 71.20

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How do you solve the triangle using law of sines given a=8, b=9, c=10?

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