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Cosine Rule

In a triangle...

Question

In a triangle ABC , A is obtuse , sinA=3/5 , sinB=5/13 then sinC=?

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Solution

sinA = 3/5 so cosA = - 4/5 (since A obtuse); similarly, using a simple right-angled triangle if cosB = 12/13, then sinB = 5/13. C = 180 - (A + B) so sinC = sin[180 - (A + B)] = sin(A + B) (4 quadrants) so sinC = sinAcosB + cosAsinB = (3/5)(12/13) + (-4/5)(5/13) = 36/65 - 20/65 = 16/65

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