Angles made by an arc

Two lamp posts and a walk

Two lamp posts $A$ and $B$ stand on the edge of a round park. You stand at a point $P$ on the edge and look at both posts. The angle between your two lines of sight is $\angle APB$. If you walk along the edge, does this angle change?

An arc $AB$ makes an angle $\angle AOB$ at the centre $O$. It makes an angle $\angle APB$ at a point $P$ on the remaining part of the circle, that is, on the other arc.

Theorem 7. The angle that an arc of a circle makes at the centre is double the angle that it makes at any point on the remaining part of the circle.

Given. A circle with centre $O$, an arc $AB$ and a point $P$ on the major arc.

To prove. $\angle AOB = 2\angle APB$.

Construction. Join $P$ to $O$ and extend it to meet the circle at $D$.

Proof.

1. In triangle $OAP$, $OA = OP$, because they are radii. So $\angle OAP = \angle OPA$. Call each one $x$.
2. The angle $AOD$ is an exterior angle of triangle $OAP$. So $\angle AOD = \angle OAP + \angle OPA = 2x$.
3. In triangle $OBP$, $OB = OP$. So $\angle OBP = \angle OPB$. Call each one $y$.
4. In the same way, $\angle BOD = 2y$.
5. Add the two: $\angle AOB = \angle AOD + \angle BOD = 2x + 2y = 2(x + y) = 2\angle APB$.

The same rule holds when the arc is a major arc. The angle at the centre is then more than $180^\circ$, and it is still double the angle at the rim.

Angles in the same segment

In the simulation, the point $P$ moves along the long arc, and the angle $APB$ never changes. This is a theorem.

Theorem 8. Angles in the same segment of a circle are equal.

Proof. Let $P$ and $Q$ be two points on the same side of the chord $AB$, on the major arc.

1. By Theorem 7, $\angle APB$ is half of $\angle AOB$.
2. By Theorem 7 again, $\angle AQB$ is also half of $\angle AOB$.
3. So $\angle APB = \angle AQB$.

This answers the question about the lamp posts. As you walk along the long arc of the park, the angle between the lines of sight stays the same.

The angle in a semicircle

Theorem 9. The angle in a semicircle is a right angle.

Proof. Let $AB$ be a diameter and $P$ a point on the circle.

1. The arc $AB$ is a semicircle, and the angle at the centre is $\angle AOB = 180^\circ$, because $A$, $O$ and $B$ lie on a straight line.
2. By Theorem 7, $\angle APB = \dfrac{180^\circ}{2} = 90^\circ$.

The converse is also true. Suppose a triangle has a right angle at $B$. Then the hypotenuse $AC$ is a diameter of the circle through the three corners. The angle at the rim is $90^\circ$, so by Theorem 7 the angle at the centre is $180^\circ$. That means $A$, $O$ and $C$ lie on one line, and the centre is the middle point of the hypotenuse.

Worked examples

Example 1. An arc $AB$ makes an angle $\angle APB = 35^\circ$ at a point $P$ on the major arc. Find $\angle AOB$ and $\angle OAB$.

1. The centre. $\angle AOB = 2 \times 35^\circ = 70^\circ$.
2. The triangle. $OA = OB$, so $\angle OAB = \dfrac{180^\circ - 70^\circ}{2} = 55^\circ$.

Example 2. $AB$ is a diameter of a circle and $P$ is a point on the circle. If $\angle PAB = 32^\circ$, find $\angle PBA$.

1. The right angle. $\angle APB = 90^\circ$, because it is the angle in a semicircle.
2. The angle sum. $\angle PBA = 180^\circ - 90^\circ - 32^\circ = 58^\circ$.

Example 3. A right triangle has sides $5\ \text{cm}$ and $12\ \text{cm}$ around the right angle. Find the radius of the circle through its three corners.

1. The hypotenuse. $\sqrt{5^2 + 12^2} = 13\ \text{cm}$.
2. The diameter. The hypotenuse is a diameter, so the radius is $6.5\ \text{cm}$.

Practise with new numbers

Try it in Python

This program fixes an arc $AB$ on a circle, then places five points on the long arc, and measures the angle at each one.

Common mistakes

• Halving the wrong way. The angle at the centre is the big one. The angle at the rim is half of it.
• Using the rule for a point on the same arc. The point $P$ must lie on the remaining part of the circle, not on the arc $AB$ itself.
• Taking the angle in a semicircle as $180^\circ$. The angle at the centre is $180^\circ$, and the angle at the rim is $90^\circ$.
• Forgetting the isosceles triangle. After finding the angle at the centre, use $OA = OB$ to find the other two angles.

Remember these

• The angle at the centre is double the angle at any point on the remaining part of the circle.
• Angles in the same segment are equal.
• The angle in a semicircle is a right angle.
• A right triangle has its hypotenuse as the diameter of the circle through its corners.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. An arc $AB$ makes an angle of $84^\circ$ at the centre. Find the angle it makes at a point on the major arc.

2. The angle at a point $P$ on the major arc is $26^\circ$. Find the angle at the centre.

3. $P$ and $Q$ are two points on the major arc $AB$. If $\angle APB = 41^\circ$, find $\angle AQB$.

4. $AB$ is a diameter and $P$ is on the circle with $\angle PBA = 64^\circ$. Find $\angle PAB$.

5. The hypotenuse of a right triangle is $20\ \text{cm}$ long. What is the radius of the circle through its corners?