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Question

Three force $$ \xrightarrow [ P ]{  } ,\xrightarrow [ Q ]{  } and\quad \xrightarrow [ R ]{  }  $$ acting with TA,IB,IC where I is the incentance of a $$ \triangle ABC $$ are in equim $$ \xrightarrow [ P ]{  } \xrightarrow [ Q ]{  } \xrightarrow [ R ]{  }  $$ is  
1113334_5163fab3b26a417cbbf2b33c53fc16de.png


Solution

According to diagram

Angle

  $$ \angle AIC=\pi -\left( \dfrac{A+C}{2} \right) $$

 $$ \angle AIC=\pi -\left( \dfrac{\pi }{2}-\dfrac{B}{2} \right) $$

 $$ \angle AIC=\dfrac{\pi }{2}+\dfrac{B}{2} $$

 Similarly,

  $$ \angle BIC=\dfrac{\pi }{2}+\dfrac{A}{2} $$

 $$ \angle AIB=\dfrac{\pi }{2}+\dfrac{C}{2} $$

Now, using Lami’s theorem

  $$ \dfrac{{\vec{P}}}{\sin \left( \dfrac{\pi }{2}+\dfrac{A}{2} \right)}=\dfrac{{\vec{Q}}}{\sin \left( \dfrac{\pi }{2}+\dfrac{B}{2} \right)}=\dfrac{{\vec{R}}}{\sin \left( \dfrac{\pi }{2}+\dfrac{C}{2} \right)} $$

 $$ \dfrac{{\vec{P}}}{\cos \dfrac{A}{2}}=\dfrac{{\vec{Q}}}{\cos \dfrac{B}{2}}=\dfrac{{\vec{R}}}{\cos \dfrac{C}{2}} $$

Hence, $$\vec{P}:\vec{Q}:\vec{R}=\cos \dfrac{A}{2}:\cos \dfrac{B}{2}:\cos \dfrac{C}{2}$$

 


Physics

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