Question

In a triangle ABC,

	Column-I		Column-II
(a)	$(c-a{)}^2=b^2-ac$ and $cosB+sinC=3/2$	(p)	$A={30}^o$
(b)	A, B and C are in A.P. and $3A=C$	(q)	$B={60}^o$
(c)	$\frac{1}{a+b}+\frac{1}{b+c}=\frac{3}{a+b+c}$	(r)	$C={90}^o$

Answer 1

(a, b)-(p, q, r); (c)-(q)
(a) Given $c^2+a^2-b^2=ac$ or $\frac{c^2+a^2-b^2}{2ac}=\frac{1}{2}$
$\therefore cosB=\frac{1}{2}\Rightarrow B={60}^o$
$\therefore sinC=\frac{3}{2}-cosB=\frac{3}{2}-\frac{1}{2}=1\therefore B={90}^o$
and hence, $A={180}^o-({90}^o+{60}^o)={30}^o$
$\therefore \left(a\right)-(p,q,r)$
(b) Given $2B=A+C={180}^o-B$
$\therefore 3B={180}^o$ or $B={60}^o$
$\therefore A+C={120}^o$ or $A+3A={120}^o$ $\therefore A={30}^o$
$\therefore C={90}^o$ $\therefore \left(b\right)-(p,q,r)$
(c) $(a+b+c)(a+b+c+b)=3(a+b)(b+c)$
or $(a+b+c{)}^2+b(a+b+c)=3(b^2+ab+bc+ca)$
$\sum a^2+2\sum ab=3\sum ab+2b^2-b(c+a)$
$a^2+c^2-b^2=\sum ab-b(c+a)=ca$
$\therefore \frac{c^2+a^2-b^2}{2ca}=\frac{1}{2}$ or $cosB=\frac{1}{2}$ $\therefore B={60}^o$
$\therefore \left(c\right)-\left(q\right)$