Circle theorems - Higher - AQAAngles at the centre and circumference - Higher

Calculate the missing angles \(x\) and \(y\).

\(x\) = \(50^\circ \times 2 = 100^\circ\)

\(y\) = \(40^\circ \times 2 = 80^\circ\)

On the diagram angle OGK = \(x\) and angle OGH = \(y\).

Angle OGK (\(x\)) = angle OKG because triangle GOK is isosceles. Lengths OK and OG are both radii.

Angle OGH (\(y\)) = angle OHG because triangle GOH is also isosceles. Lengths OH and OG are also both radii.

Angle GOH = \(180^\circ - 2y\) (because angles in a triangle add up to 180°).

Angle JOK = \(2x\) (because angles on a straight line add up to 180°).

Angle JOH = \(2y\) (because angles on a straight line add up to 180°).

The angle at the centre KOH (\(2x + 2y\)) is double the angle at the circumference KGH (\(x + y\)).