If in a triangle ABC,2cos A = sin B csc C,the | vedclass

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

Question

(C) Given the equation: $2 \cos A = \sin B \csc C$Using the sine rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,so $\frac{\sin B}{\sin C} = \frac{b}{c}$.S

Vedclass · Accepted Answer

$c = a$

Vedclass · Answer

(C) Given the equation: $2 \cos A = \sin B \csc C$Using the sine rule,$\frac{\sin B}{b} = \frac{\sin C}{c}$,so $\frac{\sin B}{\sin C} = \frac{b}{c}$.Substituting this into the given equation: $2 \cos A = \frac{b}{c}$.Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Substituting $\cos A$ into the equation: $2 \left( \frac{b^2 + c^2 - a^2}{2bc} \right) = \frac{b}{c}$.$\frac{b^2 + c^2 - a^2}{bc} = \frac{b}{c}$.Multiplying both sides by $bc$: $b^2 + c^2 - a^2 = b^2$.$c^2 - a^2 = 0 \Rightarrow c^2 = a^2$.Since $a$ and $c$ are side lengths of a triangle,$c = a$.

If in a triangle $ABC$,$2\cos A = \sin B \csc C$,then

Similar Questions

Let $ABC$ be a triangle in which $AB=BC$. Let $X$ be a point on $AB$ such that $AX:XB=AB:AX$. If $AC=AX$,then the measure of $\angle ABC$ equals (in $^{\circ}$)

If $\frac{\sin A - \sin C}{\cos C - \cos A} = \cot B$,then $A, B, C$ are in

The sines of two angles of a triangle are $\frac{5}{13}$ and $\frac{99}{101}$. The cosine of the third angle is: (in $/1313$)

Let $A, B$ and $C$ be three angles of a $\triangle ABC$ such that $\cos A + \cos B + \cos C = \frac{3}{2}$,then the $\triangle ABC$ is

The area of triangle $ABC$ is $84 \ sq. \ units$. If $AB = 13$ and $AC = 15$,then $BC$ can be ........ $units$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module