In a triangle Δ ABC,\ 3 A+4 B=6 and 4 B+3 A=1, then the angle C is

Correct Answer is : 30∘30^{}30∘

Correct Answer is : 30∘30^{}30∘

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UPSEE 2017 Solved Paper

Question : 148 of 150

Marks: +1, -0

In a triangle

\Delta ABC,\ 3\sin A+4\cos B=6

4\sin B+3\cos A=1

C

Solution:

Given,

3\sin A+4\cos B=6

.......(i)

4\sin B+3\cos A=1

.......(ii)

By squaring Eqs. (i) and (ii), we get

9\sin^{2}A+16\cos^{2}B+24\sin A\cdot\cos B

=36

......(iii)

16\sin^{2}B+9\cos^{2}A

+24\sin B\cos A=1

.......(iv)

By adding Eqs. (iii) and (iv), we get

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

+24(\sin A\cdot\cos B+\cos A\cdot\sin B)=37

⇒

9+16+24\sin(A+B)=37

[\because \sin^{2}\theta+\cos^{2}\theta=1]

\Rightarrow 24\sin(A+B)=12

\Rightarrow \sin(A+B)=\frac{1}{2}

\therefore A+B=30^{\circ}

150^{\circ}

But,

A+B=30^{\circ}

is not possible.

Because sum of two angles of a triangle is greater than third angle.

\therefore A+B=150^{\circ}

\because

Sum of angle of a triangle

=180^{\circ}

\therefore A+B+C=180^{\circ}

\Rightarrow

C=180^{\circ}-(A+B)

=180^{\circ}-150^{\circ}

=30^{\circ}

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