Question

If PQ is a tangent drawn from an external point P to a circle with centre O and QOR is a diameter where length of QOR is 8 cm such that ∠POR = 120°, then find OP and PQ.

PLEASE GIVE PROPER ANSWER!!!​

Answer 1

Answer:We are given that Circle with centre O is having diameter QR = 8 cm

Now angle PQR and angle POR is making linear pair

So angle PQR + angle POR = 180 degree angle POQ = 180 - 120

= 60

Now OR = OQ = radius = 8/2 = 4

Triangle OPQ is a right angled triangle

So sin 60 = PQ / OP

root 3 /2 = PQ / OP

PQ / root 3 = OP / 2 = K

Now OP sq. = QO sq. + PQ sq.

(2k) ^2 = ( 4) ^2 + ( root 3k)^2

4k ^2 = 16 + 3k ^2

K ^2 = 16

K = +4 & - 4

PQ = 4root 3 cm &

OP = 8 cm

I hope this helped you

Answer 2

Answer:

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