The theorem writes the equation
Sometimes the angles in a circle are not given as numbers but as expressions in x, such as 3x − 10. A circle theorem says how two angles are related, so it turns the two expressions into an equation. Solve the equation for x, then put x back into each expression to find the angles.
Every solution has the same three parts: name the theorem and write the equation it gives; solve it; check that the angles make sense.
The angle at the center
In the figure, the angle at the center AOB is (3x + 40)° and the angle at the circumference APB, on the same arc AB, is (3x − 10)°. An angle at the circumference is also called an angle at the rim.
The theorem: the angle at the center is twice the angle at the circumference. So the center expression equals 2 times the rim expression: 3x + 40 = 2(3x − 10).
Solve it. Expand the bracket: 3x + 40 = 6x − 20. Subtract 3x from both sides: 40 = 3x − 20. Add 20 to both sides: 60 = 3x. Divide by 3: x = 20.
Check. The angle at P is 3 × 20 − 10 = 50° and the angle at the center is 3 × 20 + 40 = 100°. Both are positive, and 100 = 2 × 50, as the theorem says.
The angle at the center, (3x + 40)°, and the marked angle at P, (3x − 10)°, stand on the same arc AB. With x = 20 they are 100° and 50°, and the figure is drawn at those sizes.
A cyclic quadrilateral
ABCD is a cyclic quadrilateral. The angle at A is (3x − 20)° and the angle at C, opposite it, is (2x + 30)°.
The theorem: opposite angles of a cyclic quadrilateral add to 180°. So (3x − 20) + (2x + 30) = 180.
Solve it. Collect the like terms: 5x + 10 = 180. Subtract 10 from both sides: 5x = 170. Divide by 5: x = 34.
Check. The angle at A is 3 × 34 − 20 = 82° and the angle at C is 2 × 34 + 30 = 98°. Both are positive, and 82 + 98 = 180.
The marked angles at A, (3x − 20)°, and at C, (2x + 30)°, are opposite, so they add to 180°. With x = 34 they are 82° and 98°, and the figure is drawn at those sizes.
Three facts in a row
Harder questions need more than one theorem, one after the other. ABCD is a cyclic quadrilateral in a circle with center O, and O is joined to A and to C. The angle AOC at the center, on the side of B, is (4x)°, and the angle ABC is (5x + 5)°. Find x, and then the angle OAC.
Step 1, the angle at the center. The angle ADC at the circumference stands on the same arc AC as the angle AOC, so it is half of it: ADC .
Step 2, the cyclic quadrilateral. B and D are opposite corners, so ABC + ADC = 180: (5x + 5) + 2x = 180. Then 7x + 5 = 180, so 7x = 175 and x = 25.
So the angle AOC is 4 × 25 = 100°, the angle ADC is 2 × 25 = 50°, and the angle ABC is 5 × 25 + 5 = 130°. Check: 100 = 2 × 50, and 130 + 50 = 180.
Step 3, the isosceles triangle. OA and OC are both radii, so triangle AOC is isosceles, and its two base angles are equal: OAC = OCA = (180° − 100°) ÷ 2 = 80° ÷ 2 = 40°.
Check the triangle: 100° + 40° + 40° = 180°, and every angle found is positive.
The angle at the center is marked 4x, and the marked angle at B is (5x + 5)°. With x = 25 they are 100° and 130°, and the angle at D, which stands on the same arc AC as the angle at the center, is 50°.
Why the check matters
A wrong equation usually gives an answer that fails the check. Suppose the center and rim angles 3x + 40 and 3x − 10 were set equal. Then 3x + 40 = 3x − 10 gives 40 = −10, which has no solution at all. And in the cyclic quadrilateral, setting 3x − 20 = 2x + 30 gives x = 50, which makes both angles 130°: they add to 260°, not 180°.
Angles also have to fit the figure. An angle inside a triangle, or at the corner of a quadrilateral drawn in a circle, lies between 0° and 180°. A value of x that makes one of them 0° or less, or 180° or more, means the equation was set up wrongly.
The usual mistakes
Setting the angle at the center equal to the angle at the circumference. The center angle is twice the rim angle, so the 2 goes on the rim expression: center = 2 × rim.
Putting the 2 on the wrong side. Doubling the center expression, 2(3x + 40) = 3x − 10, says the rim angle is the larger one, but the angle at the center is always the larger. Solved, it gives 6x + 80 = 3x − 10, so x = −30, and the angle at P would be 3 × (−30) − 10 = −100°, which the check rejects.
Making opposite angles add to 360°. All four angles of a quadrilateral add to 360°; one pair of opposite angles adds to 180°.
Stopping at x. When the question asks for an angle, put x back into its expression: x = 34 is not the angle at A, 82° is.
Worked example: The Lead Strips of a Round Window, Marked in x on the Plan
Question A round stained-glass window has center O. Lead strips run from O to two points A and B on the rim, and from a third point C on the rim, on the far side of the window from A and B, to A and to B. On the maker's plan, angle AOB is marked as (5x − 6)° and angle ACB is marked as (2x + 9)°. (a) Find x. (b) Find angle ACB and angle AOB.
1.Both angles stand on the arc AB. The angle at the center is twice the angle at the circumference, so 5x − 6 = 2(2x + 9).
Both angles stand on the arc AB, and the angle at the center is twice the angle at the rim: 5x − 6 = 2(2x + 9). 2.Expand the bracket: 5x − 6 = 4x + 18.
Expand the bracket: 5x − 6 = 4x + 18. 3.Subtract 4x from both sides and add 6 to both sides: x = 24. (a) x = 24.
(a) Subtract 4x and add 6: x = 24. 4.Angle ACB is 2 × 24 + 9 = 57° and angle AOB is 5 × 24 − 6 = 114°. (b) The angles are 57° at C and 114° at O.
(b) ACB = 2 × 24 + 9 = 57° and AOB = 5 × 24 − 6 = 114°. 5.Check: 114 = 2 × 57, so the angle at the center is twice the angle at the rim, and both angles are between 0° and 180°, as angles in a triangle must be.
Check: 114 = 2 × 57, the angle at the center is twice the angle at the rim.
Answer: (a) x = 24; (b) 57° and 114°
Common mistakes
- Setting the two expressions equal, 5x − 6 = 2x + 9, which gives x = 5. The two angles are not equal: the angle at the center is twice the angle at the rim, so the factor 2 must go on the angle at C.
- Putting the factor 2 on the wrong side, 2(5x − 6) = 2x + 9, which gives x = 2.625. The angle at the center is the larger one, so it is the angle at the rim that is doubled.
More congruence, similarity and circle theorems problems, worked step by step →