Mathematics
In the given figure, angle ABC = 70° and angle ACB = 50°. Given, O is the centre of the circle and PT is the tangent to the circle. Then calculate the following angles
(a) ∠CBT
(b) ∠BAT
(c) ∠PBT
(d) ∠APT
Circles
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Answer
Join AT and BT.
(a) We know that,
Angle in a semicircle is a right angle.
∴ ∠CBT = 90°.
Hence, ∠CBT = 90°.
(b) In cyclic quadrilateral ATBC,
⇒ ∠CBT + ∠CAT = 180° (∵ Sum of opposite angles of a cyclic quadrilateral = 180°)
⇒ 90° + ∠CAT = 180°
⇒ ∠CAT = 180° - 90° = 90°.
In △ABC,
⇒ ∠CBA + ∠CAB + ∠ACB = 180° [By angle sum property of triangle]
⇒ 70° + ∠CAB + 50° = 180°
⇒ ∠CAB + 120° = 180°
⇒ ∠CAB = 180° - 120°
⇒ ∠CAB = 60°.
From figure,
∠BAT = ∠CAT - ∠CAB = 90° - 60° = 30°.
Hence, ∠BAT = 30°.
(c) From figure,
∠BTX = ∠BAT = 30° [Angle in same segment are equal]
∠PBT = ∠CBT - ∠CBA = 90° - 70° = 20°.
Hence, ∠PBT = 20°.
(d) Since, ∠PTB and ∠BTX are linear pairs.
⇒ ∠PTB = 180° - ∠BTX = 180° - 30° = 150°.
In △PBT,
⇒ ∠PBT + ∠PTB + ∠APT = 180° [By angle sum property of triangle]
⇒ 20° + 150° + ∠APT = 180°
⇒ ∠APT + 170° = 180°
⇒ ∠APT = 180° - 170°
⇒ ∠APT = 10°.
Hence, ∠APT = 10°.
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