Let O be the centre of the circular ring which is suspended by the strings PA,PB,PC and PD in such a way that P is just above the point O and arcAB=arcBC=arcCD=arcAD.
∴∠AOB=90∘ Also, in △AOP, we have AP=√OA2+OP2=√32+42=5cm ⇒BA=AP=5cm In △APB, we have cosθ=