In triangle ABC,if frac{1}{a+b} + frac{1}{c+a | vedclass

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

Question

(C) Given the equation: $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Multiply both sides by $(a+b+c)$: $\frac{a+b+c}{a+b} + \frac{a+b+c}{c+a} = 3

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2}$

Vedclass · Answer

(C) Given the equation: $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Multiply both sides by $(a+b+c)$: $\frac{a+b+c}{a+b} + \frac{a+b+c}{c+a} = 3$ $1 + \frac{c}{a+b} + 1 + \frac{b}{c+a} = 3$ $\frac{c}{a+b} + \frac{b}{c+a} = 1$ $c(c+a) + b(a+b) = (a+b)(c+a)$ $c^2 + ac + ab + b^2 = ac + a^2 + bc + ab$ $b^2 + c^2 - a^2 = bc$ Using the Cosine Rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ $\cos A = \frac{bc}{2bc} = \frac{1}{2}$ Since $\cos A = \frac{1}{2}$,$A = 60^{\circ}$ Therefore,$\sin A = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$

In $\triangle ABC$,if $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\sin A$ is equal to

Similar Questions

The smallest angle of the $\Delta ABC$,when $a = 7, b = 4\sqrt{3}$ and $c = \sqrt{13}$,is .....$^o$

If the radius of the circumcircle of an isosceles triangle $PQR$ is equal to $PQ$ (where $PQ = PR$),then the angle $P$ is

Evaluate $\cot \left( \frac{A + B}{2} \right) \cdot \tan \left( \frac{A - B}{2} \right)$ in terms of sides $a$ and $b$.

The smallest angle of the triangle whose sides are $6+\sqrt{12}, \sqrt{48}, \sqrt{24}$ is

With the usual notations,in a $\triangle ABC$,if $a=2, b=\sqrt{6}$ and $c=\sqrt{3}+1$,then $\sin^2 C - \sin^2 A =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module