In the diagram below O is the center, A, C and B are points on the circumference of the circle. PCD and TBD are tangents to the circle at points C and B respectively. C1 = 36° and B1 = 62°.

The angle between the tangent PC and the chord AC is $\angle C_1 = 36^\circ$. By the Tangent-Chord Theorem, this angle is equal to the angle in the alternate segment, which is $\angle ABC$.

$$\angle ABC = \angle C_1 = 36^\circ$$

Reason: Angle between tangent and chord is equal to the angle in the alternate segment.

First, find $\angle BAC$ and $\angle ABC$.
The angle between the tangent TB and the chord BC is $\angle B_1 = 62^\circ$. By the Tangent-Chord Theorem, this angle is equal to the angle in the alternate segment, which is $\angle BAC$.

$$\angle BAC = \angle B_1 = 62^\circ$$

We already found $\angle ABC = 36^\circ$.
Next, find $\angle ACB$ using the sum of angles in $\triangle ABC$.

$$\angle ACB = 180^\circ - (\angle BAC + \angle ABC)$$ $$\angle ACB = 180^\circ - (62^\circ + 36^\circ)$$ $$\angle ACB = 180^\circ - 98^\circ = 82^\circ$$

Now, consider $\triangle ADC$.
We have $\angle DAC = \angle BAC = 62^\circ$.
We have $\angle ACD = \angle ACB = 82^\circ$.
Using the sum of angles in $\triangle ADC$:

$$\angle ADC = 180^\circ - (\angle DAC + \angle ACD)$$ $$\angle ADC = 180^\circ - (62^\circ + 82^\circ)$$ $$\angle ADC = 180^\circ - 144^\circ = 36^\circ$$

Reason: Sum of angles in a triangle.

The two other angles equal to $36^\circ$ are:

• $\boxed{\angle ABC = 36^\circ}$ (Reason: Angle between tangent and chord is equal to angle in alternate segment)
• $\boxed{\angle ADC = 36^\circ}$ (Reason: Sum of angles in $\triangle ADC$)

8.1.2 Determine with reasons the sizes of the following angles.

a. $\angle O_1$

$\angle O_1$ is $\angle AOC$. The angle subtended by arc AC at the center is $\angle AOC$, and the angle subtended by the same arc at the circumference is $\angle ABC$.
From 8.1.1, we know $\angle ABC = 36^\circ$.

$$\angle O_1 = \angle AOC = 2 \times \angle ABC$$ $$\angle O_1 = 2 \times 36^\circ = 72^\circ$$

Reason: Angle at the center is twice the angle at the circumference.

$$\boxed{\angle O_1 = 72^\circ}$$

b. $\angle C_3$

$\angle C_3$ is $\angle OCB$. In $\triangle OBC$, OB = OC (radii), so it is an isosceles triangle.
First, find $\angle BAC$. From 8.1.1, $\angle BAC = \angle B_1 = 62^\circ$ (Angle in alternate segment).
Next, find $\angle BOC$. The angle subtended by arc BC at the center is $\angle BOC$, and at the circumference is $\angle BAC$.

$$\angle BOC = 2 \times \angle BAC$$ $$\angle BOC = 2 \times 62^\circ = 124^\circ$$

Reason: Angle at the center is twice the angle at the circumference.
Now, in isosceles $\triangle OBC$:

$$\angle OCB = \angle OBC = \frac{180^\circ - \angle BOC}{2}$$ $$\angle C_3 = \frac{180^\circ - 124^\circ}{2} = \frac{56^\circ}{2} = 28^\circ$$

Reason: Base angles of an isosceles triangle and sum of angles in a triangle.

$$\boxed{\angle C_3 = 28^\circ}$$

c. $\angle D$

$\angle D$ is $\angle ADC$. We already calculated this in 8.1.1.

$$\angle ADC = 36^\circ$$

Reason: Sum of angles in $\triangle ADC$.

$$\boxed{\angle D = 36^\circ}$$

d. $\angle ACB$

From 8.1.1, we have:
$\angle BAC = 62^\circ$ (Angle in alternate segment)
$\angle ABC = 36^\circ$ (Angle in alternate segment)

$$\angle ACB = 180^\circ - (\angle BAC + \angle ABC)$$ $$\angle ACB = 180^\circ - (62^\circ + 36^\circ)$$ $$\angle ACB = 180^\circ - 98^\circ = 82^\circ$$

Reason: Sum of angles in a triangle.

$$\boxed{\angle ACB = 82^\circ}$$

8.1.3 Give a reason why CD = BD

CD and BD are both tangents drawn from the external point D to the circle.
Reason: Tangents drawn from an external point to a circle are equal in length.

$$\boxed{\text{Tangents from an external point to a circle are equal in length.}}$$