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

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

Question

(B) Given the equation: $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Taking the common denominator on the left side: $\frac{(c+a) + (b+c)}{(b+c)(

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Given the equation: $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Taking the common denominator on the left side: $\frac{(c+a) + (b+c)}{(b+c)(c+a)} = \frac{3}{a+b+c}$ $\frac{a+b+2c}{bc + ab + c^2 + ac} = \frac{3}{a+b+c}$ Cross-multiplying: $(a+b+2c)(a+b+c) = 3(bc + ab + c^2 + ac)$ $(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$ $a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$ Simplifying by subtracting common terms from both sides: $a^2 + b^2 - c^2 = ab$ Using the Cosine Rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ Substituting $a^2 + b^2 - c^2 = ab$: $\cos C = \frac{ab}{2ab} = \frac{1}{2}$ Since $\cos C = \frac{1}{2}$,we have $\angle C = 60^{\circ}$.

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

Similar Questions

If in a triangle $ABC$,$\frac{\sin A}{4} = \frac{\sin B}{5} = \frac{\sin C}{6}$,then the value of $\cos A + \cos B + \cos C$ is equal to

In a $\triangle ABC$,$(a-b)^2 \cos^2 \frac{C}{2} + (a+b)^2 \sin^2 \frac{C}{2}$ is equal to

If,in a $\triangle ABC$,$\tan \frac{A}{2} = \frac{5}{6}$ and $\tan \frac{C}{2} = \frac{2}{5}$,then $a, b, c$ are such that :

In $\Delta ABC,$ if $a = 16, b = 24$ and $c = 20,$ then $\cos \frac{B}{2} = $

In a triangle $ABC$,let $AB = \sqrt{23}$,$BC = 3$,and $CA = 4$. Then the value of $\frac{\cot A + \cot C}{\cot B}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module