Question

In a $\Delta ABC,(a+b+c)(b+c-a)=abc,then$

• A. $a<-6$
• B. $a>6$
• C. $0<a<4$
• D. $a>4$

Answer 1

Answer: (C)

$0<a<4$

$(a+b+c)(b+c-a)=abc$
$\left[\right(b+c)+a]\left[\right(b+c)-a]=abc$
$(b+c{)}^2-a^2=abc$
$b^2+c^2+2bc-a^2=abc$
$b^2+c^2-a^2=abc-2bc$
$\frac{b^2+c^2-a^2}{2bc}$ $=\frac{a}{2}-1$ $\text{[dividing both sides by 2bc]}$
$cosA=a/2-1$ $[cosA=\frac{b^2+c^2-a^2}{2bc}]$
$since,-1<cosA<1$
$so,-1<a/2-1<1$
$0<a/2<2$
$0<a<4$