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Question
In the given figure, O is the centre of the circle. PQ is a tangent to the circle at T. Chord AB produced meets the tangent at P.
AB = 9 cm, BP = 16 cm, ∠PTB = 50° ∠OBA = 45°
Find:
- Length of PT
- ∠BAT
- ∠BOT
- ∠ABT
Solution
a.
PT2 = AP × BP
= (AB + BP) × BP
= (9 + 16) × 16
= 25 × 16
= 400
∴ PT = `sqrt(400)`
= 20 cm
b. Angles in alternate segments are equal
∴ ∠BAT = ∠PTB
= 50°
c. OT ⊥ PQ
∴ ∠OTP = 90°
`\implies` ∠OTP + ∠PTB = 90°
`\implies` ∠OTB + ∠50° = 90°
`\implies` ∠OTB = 90° – 50° = 40°
Now, In ΔBOT,
OB = OT(Radii of same circle)
∠OTB = ∠OBT = 40°
and ∠BOT + ∠OTB + ∠OBT = 180°
∠BOT + 40° + 40° = 180°
∠BOT = 180° – 80°
= 100°
d. In ΔABT,
∠ABT = ∠OBA + ∠OBT
= 45° + 40°
= 85°
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