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## S Chand Class 10 ICSE Maths Solutions Chapter 14 Circle Ex 14(f)

Question 1.

In figure, PT is a tangent to a circle. If ∠BTA = 45° and ∠PTB = 75°, find ∠ABT.

Solution:

In the figure, PQ is the tangent to the circle at T

∠BTP = 75° and ∠ATB = 45°

But ∠QTA + ∠ATB + ∠BTP = 180° (Angles of a line)

∠QTA + 45° + 75° = 180° ⇒ ∠QTA + 120° = 180°

⇒ ∠QTA = 180° – 120° = 60°

Now QTP is the tangent and TA is the chord

∴ ∠QTA = ∠ABT (Angle in the alternate segment)

⇒ ∠ABT = 60°

Question 2.

In figure, TAS is a tangent to the circle, with centre O, at the point A. If ∠OBA = 32°, find the values of x and y.

Solution:

In the figure, a circle with centre O

TAS is a tangent to the circle at A

AB, AC and BC are chords of the circle

∠OBA = 32°

In △OAB,

OA = OB (radii of the same circle)

∴ ∠OAB = ∠OBA = 32°

In △AOB,

∠AOB + ∠OAB + ∠OBA = 180° (Sum of angles of a triangle)

⇒ ∠AOB + 32° + 32° = 180°

⇒ ∠AOB + 64° = 180°

⇒ ∠AOB = 180° – 64° = 116°

Now arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle

∴ ∠AOB = 2 ∠ACB

⇒ 116° = 2y ⇒ y = \(\frac{116^{\circ}}{2}\) = 58°

But ST is the tangent and AB is the chord

∴ ∠BAS = ∠ACB

⇒ x = y ⇒ x = y = 58°

Question 3.

In figure, KLMN is a cyclic quadrilateral and PQ is a tangent to the circle at K. If LN is a diameter of the circle, ∠KLN = 30° and ∠MNL = 60°, determine

(i) ∠QKN,

(ii) ∠PKL,

(iii) ∠MLK.

Solution:

In the circle with centre O,

LN is the diameter

PQ is a tangent to the circle at K

∠KLN = 30° and ∠MNL = 60°

In △LKN,

∠LKN = 90° (Angle in a semicircle)

∠KLN = 30°

∴ ∠KNL = 90° – 30° = 60°

(i) Now PQ is tangent and KN is chord of the circle at K

∴ ∠QKN = ∠KLN = 30° (Angle in the alternate segment)

(ii) Again PQ is tangent and KL is the chord

∴ ∠PKL = ∠KNL = 60°

(iii) In cyclic quad LMNK,

∠KNM + ∠KLM = 180° (sum of opposite angles)

⇒ ∠KNL + ∠LNM + ∠MLK = 180°

⇒ 60° + 60° + ∠MLK = 180°

⇒ ∠MLK + 120° = 180°

⇒ ∠MLK = 180°- 120°

⇒ ∠MLK = 60°

Question 4.

In figure, AT is a tangent to the circle. If ∠ABC = 50° and AC = BC, find ∠BAT.

Solution:

In the figure, AT is the tangent to the circle and ZABC = 50°

AC = BC

∴ ∠BAC = ∠ABC = 50° (opposite of equal sides)

But ∠ACB + ∠BAC + ∠ABC = 180° (sum of angles of a triangle)

ZACB+ 50°+ 50°= 180°

⇒ ∠ACB + 100°= 180°

⇒ ∠ACB = 180° – 100° = 80°

∵ AT is the tangent and AB is chord

∠BAT = ∠ACB = 80° (Angles in the alternate segment)

Question 5.

In figure, O is the centre of the circumcircle of △XYZ. Tangents at X and Y intersects at T. Given, ∠XTY = 80° and ∠XOZ = 140°. Calculate ∠ZXY.

Solution:

In the figure, O is the centre of circumcircle of △XYZ.

Tangents at X and Y are drawn to meet at T

∠XTY = 80° and ∠XOZ = 140°

Join OY

XT and YT are tangents to the circle

∴ ∠XTY + ∠XOY = 180° (Angles are supplementary)

⇒ 80° + ∠XOY = 180°

⇒ ∠XOY =180° – 80° = 100°

But ∠XOY + ∠YOZ + ∠ZOX = 360° (Angles at a point)

⇒ 100° +∠YOZ + 140° = 360°

⇒ ∠YOZ + 240° = 360°

∠YOZ = 360° – 240° = 120°

Now arc YZ subtends ∠YOZ at the centre and ∠ZXY at the remaining part of the circle

∴ ∠YOZ = 2 ∠ZXY

⇒ ∠ZXY = \(\frac { 1 }{ 2 }\)∠YOZ = \(\frac { 1 }{ 2 }\) × 120° = 60°

Question 6.

In figure, O is the centre of the circle and AB is a chord of the circle. Line QBS is a tangent to the circle at B. If ∠AOB = 110°, find ∠APB and ∠ABQ.

Solution:

O is the centre of the circle, AB is the chord and QBS is the tangent to the circle at B

∠AOB = 110°

Arc AB subtends ∠AOB at the centre and ∠APB at the remaining part of the circle

∴ ∠AOB = 2 ∠APB ⇒ ∠APB = \(\frac { 1 }{ 2 }\)∠AOB

⇒ ∠APB = \(\frac { 1 }{ 2 }\) × 110° = 55°

Now QBS is the tangent and AB is the chord

∴ ∠ABQ = ∠APB = 55°

Hence ∠APB = ∠ABQ = 55°

Question 7.

In figure, AB is a diameter and AC is a chord of a circle such that ∠BAC = 30°. The tangent at C intersects AB produced at D. Prove that BC = BD.

Solution:

Given : In a circle with centre O and

AB is its diameter

At C, a tangent is drawn to the circle which

meet AB on producing at D

∠BAC = 30°

To prove : BC = BD

Construction : Join BC

Proof: CD is tangent and CB is chord

∴ ∠DCB = ∠BAC = 30° (Angles in the alternate segment)

In △ABC,

∠ACB = 90° (Angle in a semicircle)

∴ ∠BAC + ∠CBA = 90°

⇒ 30° + ∠CBA = 90°

⇒ ∠CBA = 90°- 30° = 60°

Now in △BCD

Ext. ∠CBA = ∠BCD + ∠BDC

⇒ 60° = 30° + ∠BDC ⇒ ∠BDC = 60° – 30° = 30°

∵ ∠BCD = ∠BDC = 30°

BC = BD (Sides opposite to equal angles)

Hence proved.

Question 8.

In figure, DE is a tangent to the circumcircle of △ABC at the vertex A such that DE || BC. Show that AB = AC.

Solution:

In a circle, DE is a tangent at A to the circumcircle of △ABC in which DE || BC

To prove : AB = AC

Proof : ∵ DE is the tangent and AB is the chord of the circle

∴ ∠DAB = ∠ACB ….(i) (Angles in the alternate segment)

But DE || BC

∴ ∠DAB = ∠ABC ….(ii) (Alternate angles)

From (i) and (ii)

∠ACB = ∠ABC

∴ AB=AC (Sides opposite equal angles) Hence proved.

Question 9.

In figure, two circles intersect in B and C. Lines ABD and ACE are drawn to meet the second circle at points D and E, AF is a tangent at A. Prove that AF || DE.

[Hint Join B to C]

Solution:

Given : Two circles intersect each other at B and C. Lines ABD and ACE are drawn to meet the smaller circle at D and E respectively. AF is the tangent to the first circle at A

To prove : AF || DE

Construction : Join BC

Proof: AF is the tangent and AB is the chord of the first circle

∴ ∠BAF = ∠ACB ….(i) (Angles in the alternate segment)

In cyclic quadrilateral BCED,

Ext. ∠ACB = Int. opp. ∠BDE ….(ii)

From (i) and (ii)

∠BAF = ∠BDE

But these are alternate angles

∴ AF || DE

Hence proved.

Question 10.

In the figure, CD is the tangent line at C to the circumcircle of △ABC intersecting AB produced in D. Show that △DBC ~ △DCA.

Solution:

A circumcircle of △ABC, a tangent DC is drawn at C and AB is produced to meet the tangent at D

To prove : △DBC ~ △DCA

Proof: CD is tangent and BC is the chord of the circle

∴ ∠BCD = ∠BAC (Angles in the alternate segment)

Now in △DBC and △DCA,

∠D = ∠D (common)

∠BCD = ∠BAC or ∠DAC (proved)

∴ △DBC ~ △DCA

Hence proved.