O is the point of intersection of the diagonals AC and BD of a rhombus ABCD. P, Q, R are points on OC, OB and OA respectively such that OP = 1 unit, OQ = 2 units and OR = 4 units. The angle PQR is

From \( \Large \triangle OPQ, PQ^{2}=1^{2}+2^{2} = 5 \) ...(i)



From \( \Large \triangle OQR, OR^{2}=2^{2}+4^{2}=20 \)

From equation (i) and (ii)

Therefore, \( \Large PQ^{2}+QR^{2} = 5 + 20 \)

= \( \Large 25 = PR^{2} \)

\( \Large \triangle PQR = 90 ^{\circ} \)