Question

In the given figure angle ABC = 700 and angle ACB = 500. Given, O is the centre of the circle and PT is the tangent to the circle. Then calculate the following angles.

1. ∠CBT
2. ∠BAT
3. ∠PBT
4. ∠APT

Answer 1

Given:

∠ABC = 70° and ∠ACB = 50° (Assuming the ‘00’ in the text represents degrees). 

O is the center of the circle, and CT is a diameter since it passes through O.

PT is a tangent to the circle at point T.

P, A, B are collinear they lie on a straight line.

a. Calculate ∠CBT

Since CT is a diameter, the angle subtended by it at any point on the circumference is 90° in a semicircle.

 ∠CBT = 90°

b. Calculate ∠BAT

1. In ΔCBT, we know ∠CBT = 90°.

Since ∠ABC = 70°, then:

∠ABT = ∠CBT – ∠ABC

= 90° – 70°

= 20°

2. ∠ABT and ∠ACT subtend the same arc AT.

Therefore, ∠ACT = 20°.

3. Given ∠ACB = 50°

We find ∠BCT = ∠ACB – ∠ACT

= 50° – 20°

= 30°

4. ∠BCT and ∠BAT subtend the same arc BT.

Therefore, ∠BAT = 30°.

c. Calculate ∠PBT

Since P, A, B is a straight line, ∠PBT refers to the interior angle of ΔPBT at vertex B, which is the same as ∠ABT calculated above.

∠PBT = 20°

d. Calculate ∠APT

We can find this using ΔPTA:

1. Alternate Segment Theorem: The angle between tangent PT and chord AT (∠PTA) is equal to the angle in the alternate segment (∠ABT).

∠PTA = ∠ABT = 20°.

2. Since P, A, B is a straight line, the exterior angle

∠PTA = 180° – ∠BAT

= 180° – 30°

= 150°

3. In ΔPTA, the sum of angles is 180°:

∠APT + ∠PTA + ∠PAT = 180°

∠APT + 20° + 150° = 180°

∠APT = 180° – 170°

∠APT = 10°