AMU Engineering 2020 solved paper Solved Paper

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

2AP⋅BP

=‌

52+52−(3√2)2

2×5×5

[‌ In ‌∆AOB,AB2=OA2+OB2]
⇒‌‌cos‌θ=‌

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