∠APB = 63° and ∠AOB = 126°. Neither changes while P stays on the long arc, because P is not part of the arc AB that both angles stand on. Ask students what does change ∠AOB (moving A or B).

Each step changes ∠AOB by 2° and ∠APB by 1°. The centre angle always moves twice as fast, which is the theorem seen as a rate of change. Points move in 2° steps so every angle is a whole number, never rounded.

The middle letter is the vertex: ∠ABC is the angle at B between BA and BC. Hovering over a row lights up its two lines. Students who write “angle B” in a diagram with several angles at B lose marks, so practise the three-letter form early.

The angle between the radius OT and the tangent is always 90°. Exam mark schemes want the reason in words (“a tangent meets a radius at 90°”); the card gives the wording to use, so it is worth comparing students’ versions with it.

The active board has a blue outline, and the list and the theorem buttons follow it. Two boards let you compare two theorems, or the same theorem in two positions, side by side on the projector.

∠APB = 60°, half of 120°. A quick first check that students halve rather than double.

45°, 50° and 70°. Any even centre angle can be made, because points move in 2° steps. Pushing on to 180° gives 90°, which is the angle in a semicircle.

∠AOB = 70° and the reflex angle AOB = 290°. Students who give 70° for both have not noticed there are two angles at O.

Yes: ∠AOB = 40° and ∠APB = 20°. The theorem only needs both angles to stand on the same arc. Textbook diagrams nearly always show the arrowhead, so students often doubt the theorem in this position.

∠APB = 120° and the tool marks the reflex angle AOB = 240°. The angle at the centre must stand on the same arc as P, which is now the long arc. The obtuse angle at O (120°) stands on the other arc.

It stays at 63° wherever P goes on that arc. Every position of P gives half of the same 126°. That is the next theorem, angles in the same segment, found for free.

A and B must be opposite each other, with ∠AOB = 180°, so AB is a diameter. This is the angle in a semicircle as a special case of the angle at the centre.

Always 90°. While hidden it is drawn as an ordinary arc, so the square doesn’t give the answer away; it turns into a right-angle square once revealed.

B is always opposite A, so AB stays a diameter. The tool forces this because the theorem only works for a diameter. Rotating the diameter changes nothing about the 90° angle.

P halfway round, directly above or below O (90° from A and from B). The other angles are 45° each, because the angles of the triangle add up to 180° and PA = PB.

Both show 90° at P. When the angle at the centre is a straight line (180°), half of it is 90°. The angle in a semicircle is a special case of the angle at the centre, not a separate fact to memorise.

Both are 50°. P and Q are in the same segment (the shaded one), and both angles stand on the chord AB.

No. They stay equal (50° until you move A or B). Moving A or B changes both angles together.

∠AQB = 130°. The angles are no longer equal: they add up to 180°. APBQ is now a cyclic quadrilateral and these are opposite angles.

Both grow together while P and Q stay in the larger segment. When AB is a diameter both are 90°: the angle in a semicircle again.

The marked pair, ∠APB and ∠AQB, both on chord AB. ∠PAQ and ∠PBQ are also equal, as they both stand on chord PQ (A and B are on the same side of it). Students often mark the vertically opposite angles where the lines cross instead.

∠BCD = 99°, because 81° + 99° = 180°. The two blue angles are an opposite pair.

Opposite pairs always do (81° + 99° and 65° + 115° to begin with). Neighbouring angles add to 180° only in special positions, such as when AD is parallel to BC. Watch for students who add neighbours.

A rectangle (a square is one special case), and both diagonals pass through O: each is a diameter, so each corner is an angle in a semicircle.

∠BCX = 81°, the same as ∠DAB. It is 180° − ∠BCD, and ∠BCD is 180° − ∠DAB. “The exterior angle equals the interior opposite angle” is the same theorem used twice.

Yes: 100° + 80° = 180° and 140° + 40° = 180°. The theorem needs all four corners on the circle; where O is doesn’t matter.

Both 90° (the angle in a semicircle), and they still add up to 180°. ∠DAB = 85° and ∠BCD = 95° also add up to 180°.

Always 90°. The tangent turns as T moves so that it always meets the radius at a right angle.

At the top the tangent is horizontal; on the right it is vertical. In each case it is perpendicular to the radius. Students who expect tangents to “slope” are thinking of the textbook picture.

Every point on the tangent other than T is further than 5 from O, so it is outside the circle. A line through T at any other angle gets closer to O just after T, so it passes inside the circle and has to come out again.

∠ATX = 54°. Matching angles are both green. The tangent angle sits in the segment that does not contain P, which is why it is called the alternate segment.

Neither changes: 54° and 54°. Moving P along its arc is the same segment theorem. Only moving T or A changes the angles.

Both are now 126°, and the angle at T has swapped to the other side of TA, so it is still in the segment P is not in. The tangent angle always pairs with the angle in the alternate segment.

126°, because the two angles on the tangent at T make a straight line: 180° − 54°. Picking this angle instead of ∠ATX is the commonest error with this theorem.

Both 90°: ∠TPA is an angle in a semicircle, and the tangent meets the diameter (a radius) at 90°. Three theorems agree.

ET₂ = 6.4. Tangents from the same point are equal in length, so triangle T₁ET₂ is isosceles.

The angles at O and E change but always add up to 180° (104° + 76° to begin with). The right angles at T₁ and T₂ never change, and the four angles of the kite add up to 360°.

∠T₁OT₂ = 120°. Triangle T₁ET₂ is equilateral: it is isosceles (equal tangents) with a 60° angle between the equal sides. Each tangent is 8.7, and E is exactly 10 from O, twice the radius.

ET₁ = 5.0, the same as the radius, and ∠T₁ET₂ = 90°. Use ↑ and ↓ after tapping E to change its distance one degree at a time.

Both ∠OT₁E and ∠OT₂E are 90° (tangent meets radius). Knowing one tangent length gives the other, and knowing ∠T₁OT₂ (72° here) gives ∠T₁ET₂ = 108°.

MB = 4.1. The line from O meets the chord at 90° and cuts it in half, so AM = MB.

Yes to both. Triangle OAB is always isosceles (OA and OB are radii), and the line from O to the midpoint of its base meets the base at 90°.

OM = 3.0: 3² + 4² = 5² in triangle OMA, with the right angle at M. This is the 3, 4, 5 triangle that exam questions on chords use most.

The closer the chord is to O, the longer it is. The longest chord is a diameter, 10. The tool stops just short (OM = 0.3, AM = 5.0), because when the chord passes through O there is no perpendicular to draw.

∠AOB = 110° and OM splits it into two angles of 55°. In the isosceles triangle OAB, the line of symmetry OM bisects the angle at O as well as the base.

On the left the angles are 50° and 130°, adding to 180°. A, P, B and Q are four points on a circle, so APBQ is a cyclic quadrilateral and these are opposite angles. Same theorem, different picture.

Two, at T₁ and T₂. Each is the tangent and radius theorem. The two tangents theorem is the tangent and radius theorem used twice, plus congruent triangles.

∠AOB = 108°, and the tangent angle is half of it, 54°. The tangent behaves like a chord whose second end has slid all the way to T.

Yes: 104° + 76° = 180° and 90° + 90° = 180°. So O, T₁, E and T₂ lie on a circle: the kite is a cyclic quadrilateral, and OE is a diameter of that circle.

On the right ∠APB = 80°, not 90°. Only a diameter makes a semicircle. Students often see “a triangle in a circle” and assume 90°.

The 240° is the reflex angle AOB, which stands on the same arc as P. The obtuse angle AOB (120°) stands on the other arc, so it pairs with points on the long arc, not with P.

∠ATX = ∠TPA = 54°, the angle at P, opposite the chord TA. ∠TAP is a different angle (here 63°). The matching angle is always the one at the third point of the triangle, away from the tangent.

In a parallelogram opposite angles are equal; if they also add to 180° they are both 90°, so the only cyclic parallelograms are rectangles. In fact any quadrilateral with opposite angles adding to 180° is cyclic: this is the converse.

With these positions, triangle OAP has base angles of 32° and triangle OBP has base angles of 31°, so ∠APB = 63°. Each exterior angle at O is twice a base angle (64° and 62°), and 64° + 62° = 126° = ∠AOB.

Both ∠APB and ∠AQB are half of the same ∠AOB (100°), so both are 50°. Half of the same angle is the same angle.

The two centre angles on the arcs BD make a full turn, 360°. A and C are half of each (here 162° and 198°, giving 81° and 99°), so A + C = half of 360° = 180°.

Both triangles have a right angle (at T₁ and T₂), share the hypotenuse OE, and have OT₁ = OT₂ (radii). They are congruent (RHS), so ET₁ = ET₂.

The first is impossible: the opposite angles would be 80° and 80°, adding to 160°. The second works (80° + 100° each way): it is an isosceles trapezium.

The 🖨 sheet gives the revealed angles and labels every hidden one with a letter, with answers and reasons on page 2. Good problems give one angle and ask for one two steps away.