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ICSE Class 10 Mathematics Question 25 of 28

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Question 102

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

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. Maths Competency Focused Practice Questions Class 10 Solutions.
Answer

Join AT and BT.

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. Maths Competency Focused Practice Questions Class 10 Solutions.

(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°.