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

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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More