Given, 3‌sin‌A+4‌cos‌B=6 .......(i) 4‌sin‌B+3‌cos‌A=1 .......(ii) By squaring Eqs. (i) and (ii), we get 9sin2A+16cos2B+24‌sin‌A.cos‌B=36 ......(iii) 16sin2B+9cos2A+24‌sin‌B‌cos‌A=1 .......(iv) By adding Eqs. (iii) and (iv), we get 9(sin2A+cos2A)+16(cos2B+sin2B)+24(sin‌A.cos‌B+cos‌A.sin‌B)=37 ⇒ 9+16+24‌sin(A+B)=37[∵sin2θ+cos2θ=1] ⇒24‌sin(A+B)=12 ⇒sin(A+B)=
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∴A+B=30° or 150° But, A+B=30° is not possible. Because sum of two angles of a triangle is greater than third angle. ∴A+B=150° ∵ Sum of angle of a triangle =180° ∴A+B+C=180° ⇒C=180°−(A+B) =180°−150° =30°