Question

Two angles whose measures are a & b are such that $2a-3b={60}^o$ then $\frac{4a}{5b}=?$ If they form a linear pair

• A. $0$
• B. $\frac{8}{5}$
• C. $\frac{1}{2}$
• D. $\frac{2}{3}$

Answer 1

The correct option is B $\frac{8}{5}$

$2a-3b={60}^o$
$a+b={180}^o$
$b=180-a$
$2a-3b={60}^o$
$2a-3(180-a)={60}^o$ (from (i))
$2a-540+3a={60}^o$
$5a={600}^o$
$a=\frac{{600}^o}{5}$
$b={180}^o-a$
$b={180}^o-{120}^o$
$b={60}^o$
$\therefore \frac{4a}{5b}=\frac{4\times {120}^o}{5\times {60}^o}=\frac{8}{5}$