Browse Questions

# In $\Delta ABC$ if $\large\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$ then $c$ is equal to

$(a)\;90^{\large\circ}\qquad(b)\;60^{\large\circ}\qquad(c)\;45^{\large\circ}\qquad(d)\;30^{\large\circ}$