GCSE Maths: The alternate segment theorem (Higher)
Part of the GCSE Maths circle theorems chapter (Higher). Chapter overview · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)
The alternate segment theorem
Draw a circle with a tangent PQ touching it at T, and draw a chord TA starting at the point of contact. The chord makes an angle with the tangent. The chord also cuts the circle into two segments. The segment on the other side of the chord from that angle is called the alternate segment. Pick any point B on the circle in that segment and join it to T and A. The theorem says that angle TBA is equal to the angle between the tangent and the chord.
Reason to write: “the angle between a tangent and a chord is equal to the angle in the alternate segment”. A good way to use it: (1) find the chord that starts at the point of contact; (2) decide which side of that chord the tangent angle is on; (3) the equal angle is made at a point on the other side. There are two tangent-chord angles at T (one on each side of the chord), and each equals a different angle in the opposite segment.
Why is it true? Draw the diameter TD through the centre and join D to A. Let angle QTA = x. The tangent is perpendicular to the diameter at T, so angle QTD = 90° and angle DTA = 90° − x. Angle TAD is an angle in a semicircle, so it is 90°. In triangle TAD the third angle is 180° − 90° − (90° − x) = x, so angle TDA = x. Finally, angle TBA and angle TDA are in the same segment, so angle TBA = x as well. The full proof is worked example 26.7.5.
Key ideas
- The angle between a tangent and a chord is equal to the angle in the alternate segment.
- Reason: “the angle between a tangent and a chord is equal to the angle in the alternate segment”.
- The alternate segment is on the opposite side of the chord from the tangent angle.
- Each of the two angles at the point of contact has its own partner: tangent angle QTA goes with TBA; tangent angle PTB goes with TAB.
- The angle between the tangent and a chord is half of the angle at the centre on the arc that angle faces.
Watch out
The equal angle is at the other side of the chord. Do not match the tangent angle with the angle on the same side of the chord. A good check: the two equal angles should not be next to each other.Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.
Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).
- Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.Angle between tangent TQ and chord TA = 40°
- Find the alternate segment. The alternate segment is the part of the circle on the other side of the chord TA from the tangent angle. That is where B is, so angle TBA is in the alternate segment.angle TBA is in the alternate segment of the chord TA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 40° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle at the centre on arc TA would be 2 × 40° = 80°. Both angles are half of it. ✓
Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.
- Work out angle TBA
- Work out angle TAB
- Work out angle ATB
Part (a)
Method. Use the alternate segment theorem: the angle between the tangent and a chord equals the angle in the alternate segment.
- Look at the tangent angle on the chord TA. Angle QTA is between the tangent and chord TA. The alternate segment is on the other side of TA, and it contains B.angle QTA = 35°, and the matching angle is angle TBA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 35° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TA would be 70°. ✓
Watch out. The alternate segment is the one across the chord from the tangent angle.
Part (b)
Method. Use the alternate segment theorem again, this time with the chord TB.
- Look at the tangent angle on the chord TB. Angle PTB is between the tangent and chord TB. The alternate segment is on the other side of TB, and it contains A.angle PTB = 75°, and the matching angle is angle TAB
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TAB = angle PTB = 75° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TB would be 150°. ✓
Watch out. Use the angle on the opposite side of chord TB from PTB, which is at A.
Part (c)
Method. The three angles of triangle ATB add up to 180°.
- Write the two angles we know. We found them in parts (a) and (b).Angle TBA = 35° and angle TAB = 75°
- Subtract from 180°. The angles in a triangle add up to 180°.Angle ATB = 180° − 35° − 75° = 70°
Check. Also angle PTB + angle ATB + angle QTA = 75 + 70 + 35 = 180 (angles on a straight line). ✓
Watch out. You can also find this angle from the straight line PQ.
Method. First use the alternate segment theorem to get an angle at the circumference, then double it for the angle at the centre.
- Find the angle in the alternate segment. Angle QTA is between the tangent and the chord TA.Angle TBA = 38° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Name the arc. Angle TOA is at the centre and angle TBA is at the circumference. Both stand on the arc TA.arc TA
- Use the angle at the centre. The angle at the centre is twice the angle at the circumference.Angle TOA = 2 × 38° = 76° (the angle at the centre is twice the angle at the circumference)
Check. Second route: angle OTQ = 90° (tangent and radius), so angle OTA = 90° − 38° = 52°. Triangle OTA is isosceles, so angle OAT = 52° and angle TOA = 180° − 2 × 52° = 76°. ✓
Watch out. You can also find this answer without the alternate segment theorem, using the right angle at T (see the check).
- Work out the value of x.
- Work out the size of angle QTA.
Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.
- Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.(3x + 4) = (x + 28) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.3x + 4 = x + 28
- Subtract x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.3x + 4 − x = x + 28 − x
2x + 4 = 28 - Subtract 4 from both sides. We want the number terms on the other side.2x = 28 − 4
2x = 24 - Divide both sides by 2. 2x means 2 × x, so dividing by 2 leaves x on its own.x = 24 ÷ 2 = 12
- Work out the angle. Put x into the expression for angle QTA.Angle QTA = 3 × 12 + 4 = 40°
Check. Other angle: 1 × 12 + 28 = 40°, equal to the first. ✓
Watch out. These angles are equal, they do not add up to 90° or 180°.
Method. Draw the diameter TD and join DA. Find angle TDA in two ways: using the angle sum and using the same-segment theorem.
- Name the tangent angle. Use a letter so the proof works for any size of angle.Let angle QTA = x
- Use the tangent and the diameter. TD is a diameter, so it lies along the radius at T. A tangent is perpendicular to the radius at the point of contact.angle QTD = 90°, so angle DTA = 90° − x
- Use the semicircle theorem. TD is a diameter and A is on the circle.angle TAD = 90° (angle in a semicircle is 90°)
- Use the angle sum in triangle TAD. The angles add up to 180°.angle TDA = 180° − 90° − (90° − x) = x
- Move to the point B. Angles TDA and TBA are made from the same chord TA and are on the same side of it, so they are in the same segment.angle TBA = angle TDA = x (angles in the same segment are equal)
- Conclude. angle TBA = x = angle QTA. ✓
Check. Test with x = 40°: angle DTA = 50°, angle TAD = 90°, angle TDA = 40°, and 50 + 90 + 40 = 180. ✓
Watch out. In this proof the point D is in the alternate segment. If B is elsewhere in that segment, the same-segment theorem carries the result to B.
Exercise 26G The alternate segment theorem
Non-calculatorKey words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.
Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).
- Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.Angle between tangent TQ and chord TA = 36°
- Find the alternate segment. The alternate segment is the part of the circle on the other side of the chord TA from the tangent angle. That is where B is, so angle TBA is in the alternate segment.angle TBA is in the alternate segment of the chord TA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 36° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle at the centre on arc TA would be 2 × 36° = 72°. Both angles are half of it. ✓
Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.
Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.
Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).
- Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.Angle between tangent TQ and chord TA = 32°
- Find the alternate segment. The alternate segment is the part of the circle on the other side of the chord TA from the tangent angle. That is where B is, so angle TBA is in the alternate segment.angle TBA is in the alternate segment of the chord TA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 32° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle at the centre on arc TA would be 2 × 32° = 64°. Both angles are half of it. ✓
Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.
Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.
Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).
- Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.Angle between tangent TQ and chord TA = 50°
- Find the alternate segment. The alternate segment is the part of the circle on the other side of the chord TA from the tangent angle. That is where B is, so angle TBA is in the alternate segment.angle TBA is in the alternate segment of the chord TA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 50° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle at the centre on arc TA would be 2 × 50° = 100°. Both angles are half of it. ✓
Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.
Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.
Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).
- Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.Angle TBA is in the segment on the other side of TA from angle QTA.
- Find the matching angle. The alternate segment is on the other side of the chord TA from angle QTA. That is where B is, so angle TBA is the matching angle.angle TBA is in the alternate segment
- Use the alternate segment theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle QTA = angle TBA = 28° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle at the centre on arc TA would be 2 × 28° = 56°. Both angles are half of it. ✓
Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.
Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.
Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).
- Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.Angle TBA is in the segment on the other side of TA from angle QTA.
- Find the matching angle. The alternate segment is on the other side of the chord TA from angle QTA. That is where B is, so angle TBA is the matching angle.angle TBA is in the alternate segment
- Use the alternate segment theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle QTA = angle TBA = 44° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle at the centre on arc TA would be 2 × 44° = 88°. Both angles are half of it. ✓
Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.
Key words. A chord splits the circle into two segments. The alternate segment is the one on the opposite side of the chord from the angle you started with.
Method. The angle between a tangent and a chord equals the angle in the alternate segment (the segment on the opposite side of the chord).
- Spot the tangent and the chord. PQ touches the circle at T, and TA is a chord starting at the point of contact.Angle TBA is in the segment on the other side of TA from angle QTA.
- Find the matching angle. The alternate segment is on the other side of the chord TA from angle QTA. That is where B is, so angle TBA is the matching angle.angle TBA is in the alternate segment
- Use the alternate segment theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle QTA = angle TBA = 36° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. Angle at the centre on arc TA would be 2 × 36° = 72°. Both angles are half of it. ✓
Watch out. "Alternate" means the opposite side of the chord from the tangent angle. Pick the angle on the far side.
Method. Angles on the straight tangent line add up to 180°. You do not need a circle theorem for this angle.
- Spot the three angles at T. P, T and Q are on a straight line (the tangent). The lines TA and TB split it into three angles.Angle PTB = 70°, angle BTA = ?, angle ATQ = 45°
- Use the straight line PQ. The three angles at T on the line PQ add up to 180°.Angle ATB = 180° − 45° − 70° = 65° (angles on a straight line add up to 180°)
Check. 70° + 65° + 45° = 180°. ✓
Watch out. Do not forget the angle ATB sits between the two other angles on the straight line.
Method. Angles on the straight tangent line add up to 180°. You do not need a circle theorem for this angle.
- Spot the three angles at T. P, T and Q are on a straight line (the tangent). The lines TA and TB split it into three angles.Angle PTB = 75°, angle BTA = ?, angle ATQ = 30°
- Use the straight line PQ. The three angles at T on the line PQ add up to 180°.Angle ATB = 180° − 30° − 75° = 75° (angles on a straight line add up to 180°)
Check. 75° + 75° + 30° = 180°. ✓
Watch out. Do not forget the angle ATB sits between the two other angles on the straight line.
Method. Use the alternate segment theorem: the angle between the tangent and a chord equals the angle in the alternate segment.
- Look at the tangent angle on the chord TA. Angle QTA is between the tangent and chord TA. The alternate segment is on the other side of TA, and it contains B.angle QTA = 40°, and the matching angle is angle TBA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 40° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TA would be 80°. ✓
Watch out. The alternate segment is the one across the chord from the tangent angle.
Method. Use the alternate segment theorem again, this time with the chord TB.
- Look at the tangent angle on the chord TB. Angle PTB is between the tangent and chord TB. The alternate segment is on the other side of TB, and it contains A.angle PTB = 75°, and the matching angle is angle TAB
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TAB = angle PTB = 75° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TB would be 150°. ✓
Watch out. Use the angle on the opposite side of chord TB from PTB, which is at A.
Method. The three angles of triangle ATB add up to 180°.
- Write the two angles we know. We found them in parts (a) and (b).Angle TBA = 40° and angle TAB = 75°
- Subtract from 180°. The angles in a triangle add up to 180°.Angle ATB = 180° − 40° − 75° = 65°
Check. Also angle PTB + angle ATB + angle QTA = 75 + 65 + 40 = 180 (angles on a straight line). ✓
Watch out. You can also find this angle from the straight line PQ.
Method. Use the alternate segment theorem: the angle between the tangent and a chord equals the angle in the alternate segment.
- Look at the tangent angle on the chord TA. Angle QTA is between the tangent and chord TA. The alternate segment is on the other side of TA, and it contains B.angle QTA = 45°, and the matching angle is angle TBA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 45° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TA would be 90°. ✓
Watch out. The alternate segment is the one across the chord from the tangent angle.
Method. Use the alternate segment theorem again, this time with the chord TB.
- Look at the tangent angle on the chord TB. Angle PTB is between the tangent and chord TB. The alternate segment is on the other side of TB, and it contains A.angle PTB = 70°, and the matching angle is angle TAB
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TAB = angle PTB = 70° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TB would be 140°. ✓
Watch out. Use the angle on the opposite side of chord TB from PTB, which is at A.
Method. The three angles of triangle ATB add up to 180°.
- Write the two angles we know. We found them in parts (a) and (b).Angle TBA = 45° and angle TAB = 70°
- Subtract from 180°. The angles in a triangle add up to 180°.Angle ATB = 180° − 45° − 70° = 65°
Check. Also angle PTB + angle ATB + angle QTA = 70 + 65 + 45 = 180 (angles on a straight line). ✓
Watch out. You can also find this angle from the straight line PQ.
Method. Use the alternate segment theorem: the angle between the tangent and a chord equals the angle in the alternate segment.
- Look at the tangent angle on the chord TA. Angle QTA is between the tangent and chord TA. The alternate segment is on the other side of TA, and it contains B.angle QTA = 25°, and the matching angle is angle TBA
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TBA = angle QTA = 25° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TA would be 50°. ✓
Watch out. The alternate segment is the one across the chord from the tangent angle.
Method. Use the alternate segment theorem again, this time with the chord TB.
- Look at the tangent angle on the chord TB. Angle PTB is between the tangent and chord TB. The alternate segment is on the other side of TB, and it contains A.angle PTB = 85°, and the matching angle is angle TAB
- Use the theorem. The angle between a tangent and a chord is equal to the angle in the alternate segment.Angle TAB = angle PTB = 85° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
Check. The matching angle at the centre on arc TB would be 170°. ✓
Watch out. Use the angle on the opposite side of chord TB from PTB, which is at A.
Method. The three angles of triangle ATB add up to 180°.
- Write the two angles we know. We found them in parts (a) and (b).Angle TBA = 25° and angle TAB = 85°
- Subtract from 180°. The angles in a triangle add up to 180°.Angle ATB = 180° − 25° − 85° = 70°
Check. Also angle PTB + angle ATB + angle QTA = 85 + 70 + 25 = 180 (angles on a straight line). ✓
Watch out. You can also find this angle from the straight line PQ.
Method. First use the alternate segment theorem to get an angle at the circumference, then double it for the angle at the centre.
- Find the angle in the alternate segment. Angle QTA is between the tangent and the chord TA.Angle TBA = 28° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Name the arc. Angle TOA is at the centre and angle TBA is at the circumference. Both stand on the arc TA.arc TA
- Use the angle at the centre. The angle at the centre is twice the angle at the circumference.Angle TOA = 2 × 28° = 56° (the angle at the centre is twice the angle at the circumference)
Check. Second route: angle OTQ = 90° (tangent and radius), so angle OTA = 90° − 28° = 62°. Triangle OTA is isosceles, so angle OAT = 62° and angle TOA = 180° − 2 × 62° = 56°. ✓
Watch out. You can also find this answer without the alternate segment theorem, using the right angle at T (see the check).
Method. First use the alternate segment theorem to get an angle at the circumference, then double it for the angle at the centre.
- Find the angle in the alternate segment. Angle QTA is between the tangent and the chord TA.Angle TBA = 25° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Name the arc. Angle TOA is at the centre and angle TBA is at the circumference. Both stand on the arc TA.arc TA
- Use the angle at the centre. The angle at the centre is twice the angle at the circumference.Angle TOA = 2 × 25° = 50° (the angle at the centre is twice the angle at the circumference)
Check. Second route: angle OTQ = 90° (tangent and radius), so angle OTA = 90° − 25° = 65°. Triangle OTA is isosceles, so angle OAT = 65° and angle TOA = 180° − 2 × 65° = 50°. ✓
Watch out. You can also find this answer without the alternate segment theorem, using the right angle at T (see the check).
Method. First use the alternate segment theorem to get an angle at the circumference, then double it for the angle at the centre.
- Find the angle in the alternate segment. Angle QTA is between the tangent and the chord TA.Angle TBA = 32° (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Name the arc. Angle TOA is at the centre and angle TBA is at the circumference. Both stand on the arc TA.arc TA
- Use the angle at the centre. The angle at the centre is twice the angle at the circumference.Angle TOA = 2 × 32° = 64° (the angle at the centre is twice the angle at the circumference)
Check. Second route: angle OTQ = 90° (tangent and radius), so angle OTA = 90° − 32° = 58°. Triangle OTA is isosceles, so angle OAT = 58° and angle TOA = 180° − 2 × 58° = 64°. ✓
Watch out. You can also find this answer without the alternate segment theorem, using the right angle at T (see the check).
Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.
- Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.(3x + 9) = (2x + 18) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.3x + 9 = 2x + 18
- Subtract 2x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.3x + 9 − 2x = 2x + 18 − 2x
x + 9 = 18 - Subtract 9 from both sides. We want the number terms on the other side.x = 18 − 9
x = 9 - Read off x. x = 9
- Work out the angle. Put x into the expression for angle QTA.Angle QTA = 3 × 9 + 9 = 36°
Check. Other angle: 2 × 9 + 18 = 36°, equal to the first. ✓
Watch out. These angles are equal, they do not add up to 90° or 180°.
Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.
- Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.(3x + 9) = (2x + 18) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.3x + 9 = 2x + 18
- Subtract 2x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.3x + 9 − 2x = 2x + 18 − 2x
x + 9 = 18 - Subtract 9 from both sides. We want the number terms on the other side.x = 18 − 9
x = 9 - Read off x. x = 9
- Work out the angle. Put x into the expression for angle QTA.Angle QTA = 3 × 9 + 9 = 36°
Check. Other angle: 2 × 9 + 18 = 36°, equal to the first. ✓
Watch out. These angles are equal, they do not add up to 90° or 180°.
Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.
- Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.(2x + 8) = (5x − 28) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.5x − 28 = 2x + 8 (written the other way round so that the x terms stay positive)
- Subtract 2x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.5x − 28 − 2x = 2x + 8 − 2x
3x − 28 = 8 - Add 28 to both sides. We want the number terms on the other side.3x = 8 + 28
3x = 36 - Divide both sides by 3. 3x means 3 × x, so dividing by 3 leaves x on its own.x = 36 ÷ 3 = 12
- Work out the angle. Put x into the expression for angle QTA.Angle QTA = 2 × 12 + 8 = 32°
Check. Other angle: 5 × 12 − 28 = 32°, equal to the first. ✓
Watch out. These angles are equal, they do not add up to 90° or 180°.
Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.
- Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.(2x + 8) = (5x − 28) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.5x − 28 = 2x + 8 (written the other way round so that the x terms stay positive)
- Subtract 2x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.5x − 28 − 2x = 2x + 8 − 2x
3x − 28 = 8 - Add 28 to both sides. We want the number terms on the other side.3x = 8 + 28
3x = 36 - Divide both sides by 3. 3x means 3 × x, so dividing by 3 leaves x on its own.x = 36 ÷ 3 = 12
- Work out the angle. Put x into the expression for angle QTA.Angle QTA = 2 × 12 + 8 = 32°
Check. Other angle: 5 × 12 − 28 = 32°, equal to the first. ✓
Watch out. These angles are equal, they do not add up to 90° or 180°.
Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.
- Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.(4x + 6) = (3x + 17) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.4x + 6 = 3x + 17
- Subtract 3x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.4x + 6 − 3x = 3x + 17 − 3x
x + 6 = 17 - Subtract 6 from both sides. We want the number terms on the other side.x = 17 − 6
x = 11 - Read off x. x = 11
- Work out the angle. Put x into the expression for angle QTA.Angle QTA = 4 × 11 + 6 = 50°
Check. Other angle: 3 × 11 + 17 = 50°, equal to the first. ✓
Watch out. These angles are equal, they do not add up to 90° or 180°.
Method. The alternate segment theorem says the two angles are equal, so make an equation from them and solve it.
- Spot the two equal angles. Angle QTA is between the tangent and the chord TA. Angle TBA is in the alternate segment.(4x + 6) = (3x + 17) (the angle between a tangent and a chord is equal to the angle in the alternate segment)
- Write the equation. The two expressions are equal because of the circle theorem, so we can put an equals sign between them.4x + 6 = 3x + 17
- Subtract 3x from both sides. We want all the x terms on one side. Doing the same to both sides keeps the equation balanced.4x + 6 − 3x = 3x + 17 − 3x
x + 6 = 17 - Subtract 6 from both sides. We want the number terms on the other side.x = 17 − 6
x = 11 - Read off x. x = 11
- Work out the angle. Put x into the expression for angle QTA.Angle QTA = 4 × 11 + 6 = 50°
Check. Other angle: 3 × 11 + 17 = 50°, equal to the first. ✓
Watch out. These angles are equal, they do not add up to 90° or 180°.
Method. A tangent is perpendicular to the radius at the point of contact, so angle OAU is a right angle.
- Find the right angle. Radius OA meets the tangent TU at the point of contact A.Angle OAU = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle OAU is made of angle OAB and angle BAU.Angle OAB = 90° − 38° = 52°
Check. 52° + 38° = 90°. ✓
Watch out. The 90° is between the tangent and the radius, so it contains the angle OAB plus the angle UAB.
Method. Triangle OAB has two equal sides (both radii), so it is isosceles.
- Write the base angles. OA and OB are radii, so the base angles are equal.Angle OBA = angle OAB = 52° (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the two base angles from 180°.Angle AOB = 180° − 52° − 52° = 76° (the angles in a triangle add up to 180°)
Check. Alternate segment theorem: the angle in the alternate segment is 38°, and the angle at the centre is 2 × 38° = 76°. ✓
Watch out. The apex angle of the isosceles triangle is at the centre O.
Method. A tangent is perpendicular to the radius at the point of contact, so angle OAU is a right angle.
- Find the right angle. Radius OA meets the tangent TU at the point of contact A.Angle OAU = 90° (a tangent is perpendicular to the radius at the point of contact)
- Subtract. Angle OAU is made of angle OAB and angle BAU.Angle OAB = 90° − 52° = 38°
Check. 38° + 52° = 90°. ✓
Watch out. The 90° is between the tangent and the radius, so it contains the angle OAB plus the angle UAB.
Method. Triangle OAB has two equal sides (both radii), so it is isosceles.
- Write the base angles. OA and OB are radii, so the base angles are equal.Angle OBA = angle OAB = 38° (the base angles of an isosceles triangle are equal)
- Use the angle sum. Subtract the two base angles from 180°.Angle AOB = 180° − 38° − 38° = 104° (the angles in a triangle add up to 180°)
Check. Alternate segment theorem: the angle in the alternate segment is 52°, and the angle at the centre is 2 × 52° = 104°. ✓
Watch out. The apex angle of the isosceles triangle is at the centre O.
Method. Name the chord first, then find the alternate segment.
- Name the chord. Angle PTB is between the tangent TP and the chord TB.chord = TB
- Find the alternate segment. It is on the other side of TB from the angle PTB. That is the side where A is.The equal angle is at A.
- Write the correct statement. The angle at A made by the chord TB is angle TAB.angle PTB = angle TAB. Kai wrote TBA, which is at B and is made from the chord TA. So Kai is wrong.
Check. In the figure, 75° (PTB) matches 75° (TAB), while angle TBA is 35°. ✓
Watch out. The two equal angles use the same chord (here TB), one at the point of contact and one at the other point.
Method. Show two angles of triangle TAB are equal by linking both of them to the tangent angle QTA.
- Use the parallel lines. AB is parallel to PQ and TA crosses them. Angle QTA and angle TAB are alternate angles.angle TAB = angle QTA (alternate angles between parallel lines are equal)
- Use the alternate segment theorem. Angle QTA is between the tangent and the chord TA. The angle in the alternate segment is angle TBA.angle TBA = angle QTA (alternate segment theorem)
- Compare. Both angles equal angle QTA.angle TAB = angle TBA
- Conclude. A triangle with two equal angles is isosceles, and the sides opposite the equal angles are equal.So TA = TB and triangle TAB is isosceles. ✓
Check. Test: if angle QTA = 40° then both base angles are 40° and the angle at T is 100°. 40 + 40 + 100 = 180. ✓
Watch out. Use letters or the diagram to point to the angles you are comparing.