Question

Three forces P, Q, R act along the sides BC, CA and AB of a triangle ABC. If their
resultant passes through the centroid show that
(i) P/sin A + Q/sin B + R/sin C = 0

Answer 1

Given: Three forces$P,Q,R$ act along the sides$BC,CA$ and$AB$ of a triangle$ABC.$

We have to show that$\frac{P}{sinA} +\frac{Q}{sinB} +\frac{R}{sinC}$.

$\text { Let ' } O \text { ' be the circumcentre } \Rightarrow O A=O B=O C=r\\\text { consider } \triangle \mathrm{BOC}\\\Rightarrow O D=r \cdot \cos A\\\text { Similarly } O E=r \cdot \cos B, O F=r \cdot \cos C$

$\text { Let the moment of forces about be } \mathrm{N}_{0} \text {; }\\\rightarrow \text { Moment due to resultant about } 0 \text {, will be equal to sum of moment of all force about } 0 \text {. }\\M_{0}=M_{R}=M_{p}+M_{Q}+M_{R}\\\begin{array}{l}M_{P}+M_{Q}+M_{R}=P(O D)+Q(O E)+R(O F) \\=(P \cos A+P \cos B+P \cos C) r \rightarrow(1)\end{array}$$\text { Given, resultant passes through ' } 0 \text { ', }\\\Rightarrow M_{R}=0 \rightarrow(2)$

$\text { From }(1) \&(2),\\r(P \cdot \cos A+Q \cdot \cos B+R \cdot \cos C)=0\\\text { since } r \neq 0\\\Rightarrow \text { P. } \cos A+Q \cdot \cos B+R \cdot \cos C=0$

Hence.

$\Rightarrow \text { P. } \cos A+Q \cdot \cos B+R \cdot \cos C$