In a Delta ABC,the value of frac{a cos A + b | vedclass

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

Question

(A) Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into the expression:$LHS = \frac{2R \sin A \cos A + 2R

Vedclass · Accepted Answer

$\frac{r}{R}$

Vedclass · Answer

(A) Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into the expression:$LHS = \frac{2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C}{2R \sin A + 2R \sin B + 2R \sin C}$$= \frac{\sin 2A + \sin 2B + \sin 2C}{2(\sin A + \sin B + \sin C)}$Using the identities $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$ and $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$:$LHS = \frac{4 \sin A \sin B \sin C}{2(4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2})}$$= \frac{8 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2}}{8 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}$$= \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{4R} 	imes 4 = \frac{r}{R}$.

In a $\Delta ABC$,the value of $\frac{a \cos A + b \cos B + c \cos C}{a + b + c}$ is equal to:

Similar Questions

In a triangle $ABC$,if $r_1 r_2 + r_3 r = 35$,$r_2 r_3 + r r_1 = 63$,and $r_3 r_1 + r r_2 = 45$,then $2s =$

With usual notations in $\triangle ABC$,if $\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$,then $a^2, b^2, c^2$ are in

The two adjacent sides of a cyclic quadrilateral are $2$ and $5$ and the angle between them is $60^o$. If the third side is $3$,the remaining fourth side is

In $\triangle ABC$,what is the value of $\cos A + \cos B + \cos C$?

In a $\Delta ABC$,if $b = 2$,$C = 60^\circ$,and $c = \sqrt{6}$,then $a =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module