In a triangle ABC,with usual notations,(a+b+c | vedclass

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

Question

(C) Given the equation $(a+b+c)(a+b-c) = 3ab$. Expanding the left side,we get $((a+b)+c)((a+b)-c) = 3ab$. This simplifies to $(a+b)^2 - c^2 = 3ab$. Ex

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(C) Given the equation $(a+b+c)(a+b-c) = 3ab$. Expanding the left side,we get $((a+b)+c)((a+b)-c) = 3ab$. This simplifies to $(a+b)^2 - c^2 = 3ab$. Expanding $(a+b)^2$,we have $a^2 + b^2 + 2ab - c^2 = 3ab$. Rearranging the terms,we get $a^2 + b^2 - c^2 = ab$. Dividing both sides by $2ab$,we get $\frac{a^2 + b^2 - c^2}{2ab} = \frac{ab}{2ab} = \frac{1}{2}$. By the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Therefore,$\cos C = \frac{1}{2}$. Since $\cos C = \frac{1}{2}$,we have $\angle C = \frac{\pi}{3}$ or $60^\circ$.

In a triangle $ABC$,with usual notations,$(a+b+c)(a+b-c)=3ab$,then $\angle C=$

Similar Questions

If the sides of a triangle are in $A.P.$,then the cotangents of its half-angles will be in

In a triangle $ABC$,with usual notations,$2 ac \sin \left(\frac{A-B+C}{2}\right)$ is equal to

The ratios of the sides of a triangle $ABC$ are $5:12:13$ and its area is $270$. Then the sides of the triangle are:

In $\triangle ABC$,$(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to

In a $\triangle ABC$,if $(a+b+c)(b+c-a) = \lambda bc$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module