In triangle ABC,what is the value of r_1+r_2+ | vedclass

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Question

(A) We know that $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$,$r_3 = 4R

Vedclass · Accepted Answer

$4 R$

Vedclass · Answer

(A) We know that $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$,$r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$,and $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.Substituting these into the expression $r_1+r_2+r_3-r$:$= 4R [\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2} + \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2} + \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2} - \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}]$$= 4R [\cos \frac{C}{2} (\sin \frac{A}{2} \cos \frac{B}{2} + \cos \frac{A}{2} \sin \frac{B}{2}) + \sin \frac{C}{2} (\cos \frac{A}{2} \cos \frac{B}{2} - \sin \frac{A}{2} \sin \frac{B}{2})]$$= 4R [\cos \frac{C}{2} \sin (\frac{A+B}{2}) + \sin \frac{C}{2} \cos (\frac{A+B}{2})]$Since $A+B+C = \pi$,we have $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.$= 4R [\cos \frac{C}{2} \sin (\frac{\pi}{2} - \frac{C}{2}) + \sin \frac{C}{2} \cos (\frac{\pi}{2} - \frac{C}{2})]$$= 4R [\cos \frac{C}{2} \cos \frac{C}{2} + \sin \frac{C}{2} \sin \frac{C}{2}]$$= 4R [\cos^2 \frac{C}{2} + \sin^2 \frac{C}{2}] = 4R(1) = 4R$.

In $\triangle ABC$,what is the value of $r_1+r_2+r_3-r$?

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