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

If in a triangle $ABC$, $\sin A, \sin B, \sin C$ are in $A.P.$, then

If in a triangle $ABC$,$a=2$,$b=3$ and $c=4$,then $\tan \left(\frac{A}{2}\right) = $

In a triangle $ABC$,with usual notations,if $\tan \left(\frac{A}{2}\right) = \frac{5}{6}$ and $\tan \left(\frac{C}{2}\right) = \frac{2}{5}$,then:

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

In a triangle $ABC$,if $r_1 : r_2 = 3 : 4$ and $r_2 : r_3 = 2 : 3$,then $a : b : c =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module