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

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

Question

(C) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.By the sine rule,we know that $a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,wher

Vedclass · Accepted Answer

equilateral

Vedclass · Answer

(C) Given $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$.By the sine rule,we know that $a = k \sin A$,$b = k \sin B$,and $c = k \sin C$,where $k$ is a constant.Substituting these values into the given equation:$\frac{\cos A}{k \sin A} = \frac{\cos B}{k \sin B} = \frac{\cos C}{k \sin C}$$\Rightarrow \cot A = \cot B = \cot C$Since $A, B, C$ are angles of a triangle,$A = B = C$.Since all angles are equal,the triangle is an equilateral triangle.

In a $\triangle ABC$,if $\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then $\triangle ABC$ is

Similar Questions

In triangle $ABC$,if $C=120^{\circ}$,$c=\sqrt{19}$,and $b=3$,then $a=$

In a triangle $ABC$,$a[b \cos C - c \cos B] = $

If in a $\Delta ABC$,$\frac{\cos A}{a} = \frac{\cos B}{b} = \frac{\cos C}{c}$,then the triangle is

Evaluate $\cot \left( \frac{A + B}{2} \right) \cdot \tan \left( \frac{A - B}{2} \right)$ in terms of sides $a$ and $b$.

In a triangle $ABC$,if $r_1 = 2r_2 = 3r_3$,then $\sin A : \sin B : \sin C =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module