In triangle ABC with usual notation,frac{cos | 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 Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,w

Vedclass · Accepted Answer

$\frac{1}{8 \sqrt{3}}$ sq. units.

Vedclass · Answer

(D) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these,we get $\frac{\cos A}{2R \sin A} = \frac{\cos B}{2R \sin B} = \frac{\cos C}{2R \sin C}$,which implies $\cot A = \cot B = \cot C$.Since $A, B, C$ are angles of a triangle,$A = B = C = 60^\circ$,so the triangle is equilateral.Given $a = \frac{1}{\sqrt{6}}$,the area of an equilateral triangle is $\frac{\sqrt{3}}{4} a^2$.Area $= \frac{\sqrt{3}}{4} \left( \frac{1}{\sqrt{6}} ight)^2 = \frac{\sqrt{3}}{4} 	imes \frac{1}{6} = \frac{\sqrt{3}}{24} = \frac{1}{8 \sqrt{3}}$ sq. units.

In $\triangle ABC$ with usual notation,$\frac{\cos A}{a}=\frac{\cos B}{b}=\frac{\cos C}{c}$ and $a=\frac{1}{\sqrt{6}}$,then the area of the triangle is

Similar Questions

If ${c^2} = {a^2} + {b^2}$,then $4s(s - a)(s - b)(s - c) = $

In a $\triangle ABC$,$\sin A$ and $\sin B$ satisfy the equation $c^2 x^2 - c(a+b)x + ab = 0$. Then:

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:

If in a triangle $ABC$,angle $C$ is $45^o$,then $(1 + \cot A)(1 + \cot B) = $

$\alpha, \beta$ are the roots of the equation $\sin^2 x + b \sin x + c = 0$. If $\alpha + \beta = \frac{\pi}{2}$,then $b^2 - 1 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module