Find and , if A is the centre of the circle.

Question

Homi · Accepted Answer

We will use properties of circles and triangles to find the unknown angles.

Step 1 — Find equal sides

Point A is the center of the circle. Line segment AB is a radius of the circle. Line segment AC is also a radius of the circle. All radii of the same circle are equal. So, side AB is equal to side AC.

Diagram 1

Step 2 — Find equal angles

In triangle ABC, we know AB = AC. A triangle with two equal sides is an isosceles triangle. The angles opposite to equal sides are equal. Angle B is opposite to side AC. Angle C is opposite to side AB. So, angle B must be equal to angle C. Let us call angle B as $\mathbf{x}$ . Then angle C will also be $\mathbf{x}$ .

Step 3 — Use angle sum property

The sum of all angles in any triangle is 180 degrees. In triangle ABC, the angles are $\angle\text{A}$ , $\angle\text{B}$ , and $\angle\text{C}$ . We are given that $\angle\text{A}$ is 120 degrees. We found that $\angle\text{B}$ is $\mathbf{x}$ and $\angle\text{C}$ is $\mathbf{x}$ . Let us add these angles together.

$120^\circ + x + x = 180^\circ$

Let us combine the $\mathbf{x}$ terms.

$120^\circ + 2x = 180^\circ$

Now, let us subtract 120 degrees from both sides.

$2x = 180^\circ - 120^\circ$

$2x = 60^\circ$

Let us divide both sides by 2.

$x = \frac{60^\circ}{2}$

$\boxed{x = 30^\circ}$

So, angle B is $\mathbf{30^\circ}$ and angle C is $\mathbf{30^\circ}$ .

Answer

(i) $\angle\text{B} = \mathbf{30^\circ}$ (ii) $\angle\text{C} = \mathbf{30^\circ}$

Question 18

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