A-B-C-180-0-3sin-A-4cosB-6-3cosA-4sin-B-13-sin-C-

Question Number 32841 by alimudinade@gmail.com last updated on 03/Apr/18

A + B + C = 180^{0}

3 s i n A + 4 c o s B = 6

3 c o s A + 4 s i n B = \sqrt{13}

s i n C = \dots .

Answered by hknkrc46 last updated on 04/Apr/18

36 + 13 = 49

{(3 sinA + 4 cosB)}^{2} + {(3 cosA + 4 sinB)}^{2} = 49

9 (\sin^{2} A + \cos^{2} A) + 16 (\sin^{2} B + \cos^{2} B) + 24 \sin (A + B) = 49

9 + 16 + 24 \sin (180 - C) = 49

24 sinC = 24

sinC = 1

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