Question

In a $△ABC$ , if 3 < A=4 < B=6 < C,then < A , < B and < C respectively , are

• A. ${70}^0,{70}^0,{40}^0$
• B. ${80}^0,{60}^0,{40}^0$
• C. ${60}^0,{60}^0,{60}^0$
• D. ${75}^0,{45}^0,{60}^0$

Answer 1

Answer: (B)

${80}^0,{60}^0,{40}^0$

$\Rightarrow$ It is given, In $△ABC$, $3\angle A=4\angle B=6\angle C$.

$\Rightarrow$ Let $x=3\angle A=4\angle B=6\angle C$

$\Rightarrow$ $x=3\angle A$

$\therefore$ $\angle A=\frac{x}{3}$ --- ( 1 )

$\Rightarrow$ $x=4\angle B$

$\therefore$ $\angle B=\frac{x}{4}$ ---- ( 2 )

$\Rightarrow$ $x=6\angle C$

$\therefore$ $\angle C=\frac{x}{6}$ ------ ( 3 )

$\Rightarrow$ $\angle A+\angle B+\angle C={180}^o$

$\Rightarrow$ $\frac{x}{3}+\frac{x}{4}+\frac{x}{6}={180}^o$ [ From ( 1 ), ( 2 ) and ( 3 ) ]

$\Rightarrow$ $\frac{4x+3x+2x}{12}={180}^o$

$\Rightarrow$ $9x=12\times {180}^o$

$\Rightarrow$ $x=\frac{12\times 180}{9}$

$\Rightarrow$ $x={240}^o$

$\therefore$ $\angle A=\frac{x}{3}=\frac{{240}^o}{3}={80}^o$

$\therefore$ $\angle B=\frac{x}{4}=\frac{{240}^o}{4}={60}^o$

$\therefore$ $\angle C=\frac{x}{6}=\frac{{240}^o}{6}={40}^o$