I wanted to solve the following problem.In △ABC we havesin2A+sin2B=sin2C+sin⁡Asin⁡Bsin⁡CCompute sin⁡CSince it's an equation for a triangle, I assumed that π=A+B+C would be important to consider.I've tried solving for sin⁡C as a quadratic, rewriting sin⁡C=sin⁡(π−A−B), but nothing seemed to work.How does one approach this problem? Any help would be appreciated.

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I wanted to solve the following problem.In △ABC we havesin2A+sin2B=sin2C+sin⁡Asin⁡Bsin⁡CCompute sin⁡CSince it&#039;s an equation for a triangle, I assumed that π=A+B+C would be important to consider.I&#039;ve tried solving for sin⁡C as a quadratic, rewriting sin⁡C=sin⁡(π−A−B), but nothing seemed to work.How does one approach this problem? Any help would be appreciated.

muthe2ulj · Accepted Answer

using the Theorem of sines we getsin2(C)(a2+b2c2)=sin2(C)+abc2sin3(C)sincesin⁡(C)≠0we obtainsin⁡(C)=a2+b2−c2abusing2cos⁡(C)=a2+b2−c2abwe gettan⁡(C)=2

blogspainmarax6qi · Accepted Answer

By the law of sines the triangle ABC is a similar to the triangle with sides-lengths sin⁡A, sin⁡B  and  sin⁡CThus, by the law of cosinessin2C=sin2A+sin2B−2sin⁡{A}sin⁡{B}cos⁡{C}and by the givensin2C=sin2A+sin2B−sin⁡{A}sin⁡{B}sin⁡{C}Id est,2cos⁡{C}=sin⁡{C}or4(1−sin2C)=sin2Corsin⁡{C}=25orC=arcsin⁡25Done!

I wanted to solve the following problem. In

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