In a triangle ABC,if angle A = 60^{circ},then | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Apply the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.Given $\angle A = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2}$.$\frac{1}{2} = \fr

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Apply the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.Given $\angle A = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2}$.$\frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$ $\Rightarrow bc = b^2+c^2-a^2$ $\Rightarrow b^2+c^2 = bc+a^2 \dots (i)$.Now,consider the expression $\frac{b}{c+a} + \frac{c}{a+b} = \frac{b(a+b) + c(c+a)}{(c+a)(a+b)}$.$= \frac{ab+b^2+c^2+ac}{ac+a^2+bc+ab}$.Substitute $b^2+c^2 = bc+a^2$ from $(i)$:$= \frac{ab+(bc+a^2)+ac}{ac+a^2+bc+ab} = \frac{ab+bc+a^2+ac}{ab+bc+a^2+ac} = 1$.

In a $\triangle ABC$,if $\angle A = 60^{\circ}$,then $\frac{b}{c+a} + \frac{c}{a+b} = $

Similar Questions

If $A = 30^{\circ}$,$a = 7$,and $b = 8$ in $\Delta ABC$,then $B$ has

In a $\triangle ABC$,if $a=13, b=14$ and $c=15$,then $\sin(\frac{A}{2})$ is:

If the angles of a triangle are in the ratio $4:1:1$,then the ratio of the longest side to its perimeter is

In a $\triangle ABC$,if $\angle A = 3\angle B$,$CA = 9$,and $BC = 16$,then the length of $AB$ is

In $\triangle ABC$,if $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$,then $2r_1=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module