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

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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:

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