UPSEE 2017 Solved Paper

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)=

1

2

∴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°