#### Question

PT is a tangent to the circle at T. if ∠ABC = 70° and ∠ACB = 50°; Calculate:

(i) `∠`CBT (ii) `∠`BAT (iii) `∠`APT

#### Solution 1

Join AT and BT.

i) TC is the diameter of the circle

∴ `∠`CBT = 90° (Angle in a semi – circle)

(ii) `∠` CBA = 70°

∴ `∠`ABT = `∠`CBT - `∠`CBA = 90° - 70 ° = 20°

Now, `∠` ACT = `∠`ABT = 20° (Angle in the same segment of the circle)

∴`∠` TCB = `∠`ACB - `∠`ACT = 50° - 20° = 30°

But, `∠` TCB = `∠`TAB (Angles in the same segment of the circle)

∴ `∠` TAB or `∠` BAT = 30°

(iii) `∠`BTX = `∠`TCB = 30 °(Angles in the same segment)

∴ `∠` PTB = 180° - 30° =150°

Now in Δ PTB

`∠` APT + `∠`PTB +`∠` ABT = 180°

⇒ `∠`APT + 150 ° + 20° = 180°

⇒ `∠` APT = 180° - 170° = 10°

#### Solution 2

Join AT and BT.

i) TC is the diameter of the circle

∴ `∠`CBT = 90° (Angle in a semi – circle)

(ii) `∠` CBA = 70°

∴ `∠`ABT = `∠`CBT - `∠`CBA = 90° - 70 ° = 20°

Now, `∠` ACT = `∠`ABT = 20° (Angle in the same segment of the circle)

∴`∠` TCB = `∠`ACB - `∠`ACT = 50° - 20° = 30°

But, `∠` TCB = `∠`TAB (Angles in the same segment of the circle)

∴ `∠` TAB or `∠` BAT = 30°

(iii) `∠`BTX = `∠`TCB = 30 °(Angles in the same segment)

∴ `∠` PTB = 180° - 30° =150°

Now in Δ PTB

`∠` APT + `∠`PTB +`∠` ABT = 180°

⇒ `∠`APT + 150 ° + 20° = 180°

⇒ `∠` APT = 180° - 170° = 10°