Question

Two force $5kg-wt$. And $10kg-wt$ are acting with an inclination of ${120}^o$ between them. Find the angle when the resultant marks with $10kg-wt$.

Answer 1

we can use formula , $\tan \beta =\frac{B\sin \theta }{A+B\cos \theta }$

$=\frac{120\sin {120}^0}{10+5cos{120}^2}=\frac{5\times \sqrt{3}/2}{10-2.5}$

$=1/\sqrt{3}$

this gives , $\beta ={30}^0$