In any triangle ABC,with usual notations,c(a | vedclass

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

Question

(A) Using the projection formula or the law of cosines:$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$Substituting t

Vedclass · Accepted Answer

$a^2 - b^2$

Vedclass · Answer

(A) Using the projection formula or the law of cosines:$\cos B = \frac{c^2 + a^2 - b^2}{2ac}$ and $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$Substituting these into the expression:$c(a \cos B - b \cos A) = c \left( a \left( \frac{c^2 + a^2 - b^2}{2ac} \right) - b \left( \frac{b^2 + c^2 - a^2}{2bc} \right) \right)$$= c \left( \frac{c^2 + a^2 - b^2}{2c} - \frac{b^2 + c^2 - a^2}{2c} \right)$$= \frac{c^2 + a^2 - b^2 - b^2 - c^2 + a^2}{2}$$= \frac{2a^2 - 2b^2}{2} = a^2 - b^2$

In any $\triangle ABC$,with usual notations,$c(a \cos B - b \cos A) =$

Similar Questions

In a $\triangle ABC$,$b:c = \sqrt{3}:\sqrt{2}$ and the angles $A, B, C$ are in $AP$,then $\angle A = $ (in $^{\circ}$)

In $\Delta ABC,$ if $(a + b + c)(a - b + c) = 3ac,$ then

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

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module