The triangle of OQR would be an equilateral triangle since OQ = OR as they are both radii and is OQR and ORQ are 60º, ROQ is also 60º, making the triangle equilateral. Because tangents meet a circle at 90º, the angle RQP would be 30º and since angles on a straight line add to 180º, the angle QRP would be 180-ORQ which is 120º. Now we know the 2 angles in triangle of QRP so we can work out the final angle since angles in a triangle add to 180º. So:
180 - 120 - 30 = 30º.
x = 30º
This would also prove that QR and RP are the same since base angles in an isosceles triangle are equal.
