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=‌
AP2+BP2−AB2
2APâ‹…BP
=‌
52+52−(3√2)2
2×5×5
[‌ In ‌∆AOB,AB2=OA2+OB2] ⇒‌‌cos‌θ=‌