triangle ABC માં,r_1+r_2+r_3-r નું મૂલ્ય શું | vedclass

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(A) આપણે જાણીએ છીએ કે $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$,$r_3

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$4 R$

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(A) આપણે જાણીએ છીએ કે $r_1 = 4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,$r_2 = 4R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$,$r_3 = 4R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$,અને $r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.આ કિંમતોને $r_1+r_2+r_3-r$ માં મૂકતા:$= 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})]$કારણ કે $A+B+C = \pi$,તેથી $\frac{A+B}{2} = \frac{\pi}{2} - \frac{C}{2}$.$= 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$.

$\triangle ABC$ માં,$r_1+r_2+r_3-r$ નું મૂલ્ય શું છે?

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