If in a triangle ABC,frac{2 cos A}{a} + frac{ | vedclass

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

Question

(D) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting th

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(D) Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.Substituting these into the given equation:$\frac{2(b^2+c^2-a^2)}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{2(a^2+b^2-c^2)}{2abc} = \frac{a^2+b^2}{abc}$Multiplying by $2abc$:$2(b^2+c^2-a^2) + (a^2+c^2-b^2) + 2(a^2+b^2-c^2) = 2(a^2+b^2)$$2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 = 2a^2 + 2b^2$$3b^2 + c^2 + a^2 = 2a^2 + 2b^2$$b^2 + c^2 = a^2$Since $b^2 + c^2 = a^2$,by the converse of the Pythagorean theorem,$\angle A = 90^{\circ}$ or $\frac{\pi}{2}$.

If in a triangle $ABC$,$\frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$,then the value of the angle $A$ is:

Similar Questions

In a $\Delta ABC$,if $b + c = 3a$,then the value of $cot\, \frac{B}{2} \cdot cot\, \frac{C}{2}$ is equal to:

If in a $\triangle ABC$,$r_1=2$,$r_2=3$ and $r_3=6$,then $a$ equals to

In $\triangle ABC, (r_2 + r_3) \cot \left(\frac{B+C}{2}\right) = $

In a triangle $ABC$,$a = 4$,$b = 3$,and $\angle A = 60^\circ$. Then $c$ is the root of the equation:

If two angles of $\triangle ABC$ are $45^{\circ}$ and $60^{\circ}$,then the ratio of the smallest side to the greatest side is

Vedclass Test Series

Exam Paper Generator

Online Exam Module