In triangle ABC,if frac{cos A}{a}=frac{cos B} | vedclass

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

Question

(D) Given $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$.Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so $\frac{\cos A}{a} = \frac{b

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(D) Given $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$.Using the cosine rule,$\cos A = \frac{b^2+c^2-a^2}{2bc}$,so $\frac{\cos A}{a} = \frac{b^2+c^2-a^2}{2abc}$.Thus,$\frac{b^2+c^2-a^2}{2abc} = \frac{a^2+c^2-b^2}{2abc} = \frac{a^2+b^2-c^2}{2abc}$.This implies $b^2+c^2-a^2 = a^2+c^2-b^2 = a^2+b^2-c^2$.From $b^2+c^2-a^2 = a^2+c^2-b^2$,we get $2b^2 = 2a^2$,so $a=b$.From $a^2+c^2-b^2 = a^2+b^2-c^2$,we get $2c^2 = 2b^2$,so $b=c$.Therefore,$a=b=c$,which means $	riangle ABC$ is an equilateral triangle.Given $a=2$,the area is $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} (2)^2 = \sqrt{3}$ sq. units.

In $\triangle ABC$,if $\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and side $a=2$,then the area of the $\triangle ABC$ (in sq. units) is

Similar Questions

In $\Delta ABC$,the expression $a(\cos^2 B + \cos^2 C) + \cos A(c \cos C + b \cos B)$ is equal to:

If $\alpha$ and $\beta$ are different values of $x$ satisfying $a \cos x + b \sin x = c,$ then $\tan \left( \frac{\alpha + \beta}{2} \right) = $

The equation $\sin ^4 x-(k+3) \sin ^2 x-k-4=0$ has a solution if

Let $S=\{\theta \in[0,2 \pi): \tan (\pi \cos \theta)+\tan (\pi \sin \theta)=0\}$. Then $\sum_{\theta \in S } \sin ^2\left(\theta+\frac{\pi}{4}\right)$ is equal to

If triangle $ABC$ is right-angled at $A$ and $\tan \frac{B}{2}, \tan \frac{C}{2}$ are roots of the equation $ax^2 + bx + c = 0$ $(a \neq 0)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module