Tangents & Intersecting Chords — Test Yourself

AP is a tangent to the given circle. If AB = 8 cm and BC = 10 cm, then AP is :

1. 8 cm

2. 16 cm

3. 12 cm

4. 24 cm

Answer

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

From figure,

AC = AB + BC = 8 + 10 = 18 cm.

⇒ AB × AC = AP²

⇒ 8 × 18 = AP²

⇒ AP² = 144

⇒ AP = $\sqrt{144}$ = 12 cm.

Hence, Option 3 is the correct option.

In the given figure, O is center of the circle and PQ is a tangent. If angle OAB = x; the measure of angle ABP; in terms of x, is :

1. x

2. 180° - 2x

3. 90° + x

4. 90° - x

Answer

From figure,

In △OAB,

OA = OB (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

⇒ ∠OBA = ∠OAB = x

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ OB ⊥ PQ

∴ ∠PBO = 90°

From figure,

∠ABP = ∠PBO - ∠OBA = 90° - x.

Hence, Option 4 is the correct option.

In the given figure, AB is tangent to the circle with center O. If OCB is a straight line segment, the angle BAC is :

1. 40°

2. 55°

3. 35°

4. 20°

Answer

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ OA ⊥ AB

∴ ∠OAB = 90°.

Let, ∠BAC = x

From figure,

In △OAC,

∠A = ∠OAB - ∠BAC = 90° - x.

Also,

OA = OC (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠C = ∠A = 90° - x.

By angle sum property of triangle,

⇒ ∠A + ∠O + ∠C = 180°

⇒ 90° - x + ∠O + 90° - x = 180°

⇒ ∠O + 180° - 2x = 180°

⇒ ∠O = 180° - 180° + 2x = 2x.

In △OAB,

By angle sum property of triangle,

⇒ ∠O + ∠A + ∠B = 180°

⇒ ∠O + ∠OAB + ∠B = 180°

⇒ 2x + 90° + 20° = 180°

⇒ 2x + 110° = 180°

⇒ 2x = 180° - 110°

⇒ 2x = 70°

⇒ x = $\dfrac{70°}{2}$ = 35°.

⇒ ∠BAC = 35°.

Hence, Option 3 is the correct option.

In the given figure O is center, PQ is tangent at point A. BD is diameter and ∠AOD = 84° then angle QAD is :

1. 32°

2. 84°

3. 48°

4. 42°

Answer

In △OAD,

OA = OD (Radius of same circle)

We know that,

Angles opposite to equal sides are equal.

∴ ∠A = ∠D = x (let)

⇒ ∠O + ∠A + ∠D = 180° (By angle sum property of triangle)

⇒ 84° + x + x = 180°

⇒ 2x = 180° - 84°

⇒ 2x = 96°

⇒ x = $\dfrac{96°}{2}$ = 48°.

From figure,

∠OAD = ∠A = 48°

We know that,

Tangent at any point of a circle and the radius through this point are perpendicular to each other.

∴ ∠OAQ = 90°

From figure,

∠DAQ = ∠OAQ - ∠OAD = 90° - 48° = 42°.

Hence, Option 4 is the correct option.

Two mutually perpendicular tangents are drawn to a circle with radius $R\sqrt{2}$ units. The shortest distance between the two points of contact is :

1. R units

2. $\dfrac{1}{2}R$ units

3. $R\sqrt{2}$ units

4. 2R units

Answer

Let two perpendicular tangents from external point A touch the circle at points B and C.

Given,

Radius = $R\sqrt{2}$ units

From figure,

AC = OB = $R\sqrt{2}$,

AB = OC = $R\sqrt{2}$.

In right angle triangle ABC,

⇒ BC² = AB² + AC²

⇒ BC² = $(R\sqrt{2})^2 + (R\sqrt{2})^2$

⇒ BC² = 2R² + 2R²

⇒ BC² = 4R²

⇒ BC = $\sqrt{4R^2}$

⇒ BC = 2R units.

Hence, Option 4 is the correct option.

Three circles with centers A, B and C and radii 5 cm, 2 cm and 6 cm respectively and touch each other externally.

Assertion (A): To find the perimeter of the triangle ABC, add the radii of given three circles.

Reason (R): The required perimeter is the product of sum of the radii and 2.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

Let the circles intersect at points D, E and F.

From figure,

Perimeter of triangle ABC = AB + BC + CA

= (AD + BD) + (BE + CE) + (CF + FA)

= 5 + 2 + 2 + 6 + 6 + 5

= 26 cm.

On adding radii of three circles, we get :

5 + 2 + 6 = 13 cm, which is not equal to perimeter.

Sum of radii × 2 = 13 × 2 = 26 cm, which is equal to perimeter.

∴ A is false, R is true

Hence, Option 2 is correct option.

AB is diameter of the circle. PA is tangent and ∠AOC = 60°.

Assertion(A): x + 30° = 90°.

Reason(R): PA is tangent

⇒ ∠BAP = 90°

∴ x + 30° = 90°

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

We know that,

The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠AOC = 2 x ∠ABC

⇒ 60° = 2 x ∠ABC

⇒ ∠ABC = $\dfrac{60°}{2}$ = 30°

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ AP ⊥ OA

⇒ ∠OAP = 90°

⇒ ∠OAP = ∠BAP = 90°

In ΔABP, according to angle sum property,

∴ ∠ABP + ∠APB + ∠BAP = 180°

⇒ 30° + x + 90° = 180°

⇒ 30° + x = 180° - 90°

⇒ 30° + x = 90°.

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Chords AD and BC one produce meet at exterior point P.

Assertion(A): PD x AD = PC x BC.

Reason(R): In triangles PAB and PCD.

∠PAB = ∠PCD ⇒ ΔPAB ∼ ΔPCD

$\Rightarrow \dfrac{PD}{PB} = \dfrac{PC}{PA}$

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is correct reason for A.

4. Both A and R are true and R is incorrect reason for A.

Answer

In ΔPAB and ΔPCD,

⇒ ∠PBA = ∠PDC (Exterior angles of a cyclic quadrilateral is always equal to interior opposite angle)

⇒ ∠PAB = ∠PCD ((Exterior angles of a cyclic quadrilateral is always equal to interior opposite angle))

ΔPAB ~ ΔPCD (By A.A. axiom of similarity)

We know that,

Corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{PD}{PB} = \dfrac{PC}{PA}$

∴ Reason (R) is true.

⇒ PD x PA = PC x PB

∴ Assertion (A) is false.

∴ A is false, R is true.

Hence, option 2 is the correct option.

Two circles touch each other externally at point P. OA and OB are the tangent of the two circles (as shown) and OA = 10 cm.

Statement (1): OB = 10 cm.

Statement (2): On joining O and P, tangent OP = tangent OA and tangent OP = tangent OB

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Join OP.

We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From figure,

O is the point from which, OA and OP are two tangents to the circle with centre Q'.

So, OA = OP .......(1)

Similarly, from point O, OB and OP are two tangents to the circle with centre Q.

So, OB = OP ......(2)

From (1) and (2), we have

⇒ OA = OB

⇒ OB = 10 cm

∴ Both the statements are true.

Hence, option 1 is the correct option.

O is centre of the circle, PB and PC are tangents and ∠BPC = 50°.

Statement (1): ∠BAC = ∠P = 50°

Statement (2): ∠BOC + 50° = 180°

⇒ ∠BOC = 130°

∴ ∠BAC = 65°

1. Both the statement are true.

2. Both the statement are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ OB ⊥ BP and OC ⊥ CP

⇒ ∠OBP = 90° and ∠OCP = 90°

OCPB is a quadrilateral.

∴ ∠OBP + ∠BPC + ∠OCP + ∠BOC = 360°

⇒ 90° + 50° + 90° + ∠BOC = 360°

⇒ 230° + ∠BOC = 360°

⇒ ∠BOC = 360° - 230°

⇒ ∠BOC = 130°

We know that,

The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BOC = 2 x ∠BAC

⇒ 130° = 2 x ∠BAC

⇒ ∠BAC = $\dfrac{130°}{2}$ = 65°.

So, Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

In the given figure, C and D are points on the semi-circle described on AB as diameter.

Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC.

Answer

As ABCD is a cyclic quadrilateral, we have

⇒ ∠BCD + ∠BAD = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

⇒ ∠BCD + 70° = 180°

⇒ ∠BCD = 180° - 70° = 110°

In ∆BCD,

⇒ ∠CBD + ∠BCD + ∠BDC = 180° [By angle sum property of triangle]

⇒ 30° + 110° + ∠BDC = 180°

⇒ ∠BDC = 180° - 140° = 40°.

Hence, ∠BDC = 40°.

In cyclic quadrilateral ABCD, ∠A = 3∠C and ∠D = 5∠B. Find the measure of each angle of the quadrilateral.

Answer

Given, cyclic quadrilateral ABCD

So, ∠A + ∠C = 180° [Opposite angles in a cyclic quadrilateral is supplementary]

⇒ 3∠C + ∠C = 180° [As ∠A = 3∠C]

⇒ 4∠C = 180°

⇒ ∠C = $\dfrac{180°}{4}$

⇒ ∠C = 45°.

Now,

⇒ ∠A = 3∠C = 3 x 45° = 135°.

Similarly,

⇒ ∠B + ∠D = 180°

⇒ ∠B + 5∠B = 180° [As, ∠D = 5∠B]

⇒ 6∠B = 180°

⇒ ∠B = $\dfrac{180°}{6}$

⇒ ∠B = 30°.

Now,

⇒ ∠D = 5∠B = 5 x 30° = 150°.

Hence, ∠A = 135°, ∠B = 30°, ∠C = 45° and ∠D = 150°.

Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Answer

Let circle be drawn on one of the equal sides AB of the isosceles triangle ABC as shown in the figure below:

We have ∠ADB = 90° [Angle in a semi-circle is a right angle]

But,

⇒ ∠ADB + ∠ADC = 180° [Linear pair]

⇒ 90° + ∠ADC = 180°

⇒ ∠ADC = 180° - 90°

⇒ ∠ADC = 90°.

In ∆ABD and ∆ACD, we have

⇒ ∠ADB = ∠ADC [Each 90°]

⇒ AB = AC [Given]

⇒ AD = AD [Common]

Hence, ∆ABD ≅ ∆ACD by RHS congruence criterion.

By, C.P.C.T we get :

BD = DC

Hence, the circle bisects base BC at D.

Bisectors of vertex angles A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = 90° - $\dfrac{1}{2}$∠A.

Answer

BE is the bisector of ∠B.

⇒ ∠ABE = $\dfrac{∠B}{2}$

From figure,

⇒ ∠ADE = ∠ABE [Angles in same segment are equal]

⇒ ∠ADE = $\dfrac{∠B}{2}$ ...........(1)

Also,

FC is the bisector of ∠C.

⇒ ∠ACF = $\dfrac{∠C}{2}$

⇒ ∠ACF = ∠ADF [Angles in same segment are equal]

⇒ ∠ADF = $\dfrac{∠C}{2}$ .............(2)

From figure,

⇒ ∠D = ∠ADE + ∠ADF

⇒ ∠D = $\dfrac{∠B + ∠C}{2}$ ...........(3)

In triangle ABC,

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° - ∠A

Substituting above value in (3), we get :

⇒ ∠D = $\dfrac{180° - ∠A}{2}$

⇒ ∠D = 90° - $\dfrac{1}{2}$∠A.

From figure,

∠EDF = ∠D.

Hence, proved that ∠EDF = $90° - \dfrac{1}{2}∠A$.

In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.

Answer

Join OB.

Given,

BC = OD = radius of circle

⇒ BC = OB.

As, angles opposite to equal sides are equal.

⇒ ∠BOC = ∠BCO = 20°.

An exterior angle is equal to the sum of two opposite interior angles.

∴ ∠ABO = ∠BCO + ∠BOC = 20° + 20° = 40° ............(1)

Now in ∆OAB,

OA = OB [Radii of the same circle]

As, angles opposite to equal sides are equal.

∠OAB = ∠ABO = 40° [from (1)]

By angle sum property of triangle,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 40° + 40° = 180°

⇒ ∠AOB + 80° = 180°

⇒ ∠AOB = 180° - 80°

⇒ ∠AOB = 100°

As DOC is a straight line,

⇒ ∠AOD + ∠AOB + ∠BOC = 180°

⇒ ∠AOD + 100° + 20° = 180°

⇒ ∠AOD = 180° - 120° = 60°.

Hence, ∠AOD = 60°.

P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.

Answer

We know that,

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

As, TPS is a tangent and PA is the chord of the circle.

∠BPT = ∠PAB [Angles in alternate segments are equal] ..........(1)

But,

∠PBA = ∠PAB [Since, PA = PB as P is mid-point of arc APB.] ........(2)

From (1) and (2), we get :

∠BPT = ∠PBA

The above angles are alternate angles,

∴ TPS || AB

Hence, proved that the tangent drawn at P will be parallel to the chord AB.

In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find:

(i) ∠DBC

(ii) ∠BCP

(iii) ∠ADB

Answer

(i) We know that,

The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment, we have :

From figure,

PQ is a tangent and CD is a chord.

⇒ ∠DBC = ∠DCQ [Angles in the alternate segment are equal]

⇒ ∠DBC = 40°.

Hence, ∠DBC = 40°.

(ii) In △DCB

⇒ ∠DBC + ∠DCB + ∠CDB = 180° [By angle sum property of triangle]

⇒ 40° + 90° + ∠CDB = 180° [∠DCB = 90°, as angle in a semi-circle is a right angle]

⇒ ∠CDB = 180° - 130° = 50°.

From figure,

⇒ ∠BCP = ∠CDB = 50°. [Angles in the alternate segment are equal]

Hence, ∠BCP = 50°.

(iii) In ∆ABD,

∠BAD = 90° [Angle in a semi-circle is a right angle]

∠ABD = 60° [Given]

⇒ ∠ADB + ∠BAD + ∠ABD = 180° [By angle sum property of triangle]

⇒ ∠ADB + 90° + 60° = 180°

⇒ ∠ADB = 180° - 150° = 30°

Hence, ∠ADB = 30°.

The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that : ∠ACD + ∠BAC = 90°.

Answer

Join OC.

We know that,

The radius from the center of the circle to the point of tangent is perpendicular to the tangent line.

BCD is the tangent and OC is the radius.

As, OC ⊥ BD

∠OCD = 90°

∴ ∠OCA + ∠ACD = 90° ............. (1)

In ∆OCA,

⇒ OA = OC [Radius of the same circle]

∴ ∠OCA = ∠OAC [As, angles opposite to equal sides are equal]

Substituting in (1), we get

⇒ ∠OAC + ∠ACD = 90°

⇒ ∠BAC + ∠ACD = 90° [From figure, ∠BAC = ∠OAC]

Hence, proved that ∠ACD + ∠BAC = 90°.

ABC is a right triangle with angle B = 90°. A circle with BC as diameter meets hypotenuse AC at point D. Prove that:

(i) AC x AD = AB²

(ii) BD² = AD x DC.

Answer

(i) In ∆ABC, we have

∠B = 90° and BC is the diameter of the circle.

Hence, AB is the tangent to the circle at B.

We know that,

If a chord and a tangent intersect externally, then the product of the lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ AB² = AD x AC.

Hence, proved that AB² = AD x AC.

(ii) From figure,

∠BDC = 90° [Angle in a semi-circle is a right angle.]

From figure,

⇒ ∠ADB + ∠BDC = 180° [Linear pairs]

⇒ ∠ADB + 90° = 180°

⇒ ∠ADB = 180° - 90°

⇒ ∠ADB = 90°

In ∆ADB,

⇒ ∠ADB + ∠A + ∠ABD = 180° [By angle sum property of triangle]

⇒ 90° + ∠A + ∠ABD = 180°

⇒ ∠A + ∠ABD = 90° ...............(1)

In ∆ABC, ∠ABC = 90°.

⇒ ∠ABC + ∠A + ∠ACB = 180° [By angle sum property of triangle]

⇒ 90° + ∠A + ∠ACB = 180°

⇒ ∠A + ∠ACB = 90° ...............(2)

From (1) and (2),

⇒ ∠A + ∠ABD = ∠A + ∠ACB

⇒ ∠ABD = ∠ACB.

From figure,

⇒ ∠ACB = ∠BCD

∴ ∠ABD = ∠BCD

Now in ∆ABD and ∆CBD, we have

∠BDA = ∠BDC [Both equal to 90°]

∠ABD = ∠BCD

Hence, ∆ABD ~ ∆CBD by AA postulate.

We know that,

Ratio of corresponding sides of similar triangles are same.

$\dfrac{BD}{DC} = \dfrac{AD}{BD}$

∴ BD² = AD x DC.

Hence, proved that BD² = AD x DC.

In the given figure, AC = AE.

Show that :

(i) CP = EP

(ii) BP = DP

Answer

(i) In ∆ADC and ∆ABE,

⇒ ∠ACD = ∠AEB [Angles in the same segment are equal]

⇒ AC = AE [Given]

⇒ ∠A = ∠A [Common]

Hence, ∆ADC ≅ ∆ABE by ASA axiom.

So, by C.P.C.T we have

⇒ AD = AB ..............(1)

Given,

⇒ AE = AC .............(2)

Subtracting equation (1) from (2), we get :

⇒ AE - AD = AC - AB

⇒ DE = BC

In ∆BPC and ∆DPE,

⇒ ∠C = ∠E [Angles in the same segment are equal]

⇒ BC = DE [Proved above]

⇒ ∠CBP = ∠PDE [Angles in the same segment are equal]

Hence, ∆BPC ≅ ∆DPE by ASA axiom.

So, by C.P.C.T we have

⇒ CP = EP

Hence, proved that CP = EP.

(ii) Proved above,

∆BPC ≅ ∆DPE

∴ BP = DP [By C.P.C.T]

Hence, proved that BP = DP.

In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If ∠ACO = 30°, find:

(i) ∠BCO

(ii) ∠AOB

(iii) ∠APB

Answer

(i) In ∆OAC and ∆OBC, we have

⇒ OC = OC [Common]

⇒ OA = OB [Radii of the same circle]

⇒ CA = CB [Tangents to the circle from an exterior point are equal.]

Hence, ∆OAC ≅ ∆OBC by SSS congruence criterion.

∴ ∠ACO = ∠BCO = 30° [By C.P.C.T.]

Hence, ∠BCO = 30°.

(ii) From figure,

∠ACB = ∠ACO + ∠BCO = 30° + 30° = 60°.

Sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠AOB + ∠ACB = 180°

⇒ ∠AOB = 180° - 60° = 120°.

Hence, ∠AOB = 120°.

(iii) Arc AB subtends ∠AOB at the center and ∠APB is the remaining part of the circle.

We know that.

When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference.

⇒ ∠APB = $\dfrac{1}{2}$∠AOB = $\dfrac{1}{2}$ x 120° = 60°.

Hence, ∠APB = 60°.

The given figure shows a semi-circle with center O and diameter PQ. If PA = AB and ∠BCQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Answer

Join PB.

In cyclic quadrilateral PBCQ,

⇒ ∠BPQ + ∠BCQ = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ ∠BPQ + 140° = 180°

⇒ ∠BPQ = 180° - 140°

⇒ ∠BPQ = 40° ............(1)

In △PBQ,

∠PBQ = 90° [Angle in a semi-circle is a right angle.]

⇒ ∠PBQ + ∠BPQ + ∠PQB = 180° [Angle sum property of triangle]

⇒ 90° + 40° + ∠PQB = 180°

⇒ 130° + ∠PQB = 180°

⇒ ∠PQB = 180° - 130°

⇒ ∠PQB = 50°.

In cyclic quadrilateral PQBA,

⇒ ∠PQB + ∠PAB = 180° [Sum of opposite angles of a cyclic quadrilateral = 180°]

⇒ 50° + ∠PAB = 180°

⇒ ∠PAB = 180° - 50°

⇒ ∠PAB = 130°.

In △PAB,

⇒ ∠PAB + ∠PBA + ∠BPA = 180° [Angle sum property of triangle]

⇒ 130° + ∠PBA + ∠BPA = 180°

⇒ ∠PBA + ∠BPA = 180° - 130°

⇒ ∠PBA + ∠BPA = 50° ................(2)

Given, PA = PB

Angles opposite to equal sides are equal.

∴ ∠PBA = ∠BPA = x (let)

Substituting above value in (2), we get :

⇒ x + x = 50°

⇒ 2x = 50°

⇒ x = $\dfrac{50°}{2}$

⇒ x = 25°.

From figure,

∠AQB = ∠APB = 25° [Angles in same segment are equal.]

∠APQ = ∠APB + ∠BPQ = 25° + 40° = 65.

We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc AQ subtends ∠AOQ at the center and ∠APQ at the remaining part of the circle.

∠AOQ = 2∠APQ = 2 × 65° = 130°.

In △AOQ,

OA = OQ [Radii of same circle]

As, angles opposite to equal sides are equal.

∠OAQ = ∠OQA = y (let)

⇒ ∠OAQ + ∠OQA + ∠AOQ = 180° [Angle sum property of triangle]

⇒ y + y + 130° = 180°

⇒ 2y = 180° - 130°

⇒ 2y = 50°

⇒ y = $\dfrac{50°}{2}$

⇒ y = 25°.

∴ ∠OAQ = 25°.

Since, ∠OAQ = ∠AQB = 25°.

∠OAQ and ∠AQB are alternate angles.

Thus, AO and BQ are parallel.

Hence, ∠AQB = 25° and ∠PAB = 130°.

The given figure shows a circle with center O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate :

(i) angle QTR

(ii) angle QRP

(iii) angle QRS

(iv) angle STR

Answer

(i) From figure,

⇒ ∠POQ + ∠QOR = 180° [Linear pairs]

⇒ 100° + ∠QOR = 180°

⇒ ∠QOR = 180° - 100°

⇒ ∠QOR = 80°.

We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc RQ subtends ∠QOR at the center and ∠QTR at the remaining part of the circle.

⇒ ∠QOR = 2∠QTR

⇒ ∠QTR = $\dfrac{1}{2}$∠QOR = $\dfrac{1}{2} \times 80° = 40°.$

Hence, ∠QTR = 40°.

(ii) We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc QP subtends ∠QOP at the center and ∠QRP at the remaining part of the circle.

⇒ ∠QOP = 2∠QRP

⇒ ∠QRP = $\dfrac{1}{2}$∠QOP = $\dfrac{1}{2} \times 100° = 50°.$

Hence, ∠QRP = 50°.

(iii) Given,

RS || QT

⇒ ∠SRT = ∠QTR = 40° (Alternate angles are equal)

From figure,

∠QRS = ∠QRP + ∠PRT + ∠SRT = 50° + 20° + 40° = 110°.

Hence, ∠QRS = 110°.

(iv) Since, RSTQ is a cyclic quadrilateral and sum of opposite angles of cyclic quadrilateral = 180°.

⇒ ∠QRS + ∠QTS = 180°

⇒ ∠QRS + ∠QTR + ∠STR = 180°

⇒ 110° + 40° + ∠STR = 180°

⇒ ∠STR = 180° - 150°

⇒ ∠STR = 30°.

Hence, ∠STR = 30°.

In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.

If ∠AXB = 50° and ∠ABX = 70°, find ∠BAY and ∠APY.

Answer

In △AXB,

⇒ ∠AXB + ∠XAB + ∠ABX = 180° [Angle sum property of triangle]

⇒ 50° + XAB + 70° = 180°

⇒ ∠XAB = 180° - 120° = 60°.

From figure,

∠XAY = 90° [Angle in a semi-circle is a right angle.]

∠BAY = ∠XAY - ∠XAB = 90° - 60 = 30°.

∠BXY = ∠BAY = 30° [Angles in same segment are equal]

We know that,

An exterior angle is equal to the sum of two opposite interior angles.

⇒ ∠ACX = ∠BXC + ∠CBX

⇒ ∠ACX = ∠BXY + ∠ABX [From figure, ∠BXC = ∠BXY and ∠CBX = ∠ABX]

⇒ ∠ACX = 30° + 70° = 100°.

We know that,

Diameter is perpendicular to tangent.

⇒ ∠XYP = 90°

An exterior angle in a triangle is equal to sum of two opposite interior angles.

⇒ ∠ACX = ∠APY + ∠CYP

⇒ ∠APY = ∠ACX - ∠CYP = 100° - 90° = 10°.

Hence, ∠APY = 10° and ∠BAY = 30°.

In the given figure, QAP is the tangent at point A and PBD is a straight line.

If ∠ACB = 36° and ∠APB = 42°, find :

(i) ∠BAP

(ii) ∠ABD

(iii) ∠QAD

(iv) ∠BCD

Answer

(i) We know that,

The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.

∴ ∠BAP = ∠ACB = 36°.

Hence, ∠BAP = 36°.

(ii) We know that,

An exterior angle in a triangle is equal to the sum of two opposite interior angles.

In △APB,

∠ABD = ∠APB + ∠BAP = 42° + 36° = 78°.

Hence, ∠ABD = 78°.

(iii) From figure,

∠ADB = ∠ACB = 36° (Angles in same segment are equal)

In △PAD,

∠QAD = ∠APB + ∠ADB = 42° + 36° = 78°. [Exterior angle is equal to sum of two opposite interior angles.]

Hence, ∠QAD = 78°.

(iv) We know that,

The angle between a tangent and a chord through the point of contact is equal to an angle in the alternate segment.

∴ ∠ACD = ∠QAD = 78°.

From figure,

∠BCD = ∠ACB + ∠ACD = 36° + 78° = 114°.

Hence, ∠BCD = 114°.

In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.

If ∠CAB = 34°, find :

(i) ∠CBA

(ii) ∠CQB

Answer

(i) From figure,

∠ACB = 90° (Angle in a semi-circle is a right angle.)

In △ACB,

⇒ ∠ACB + ∠CAB + ∠CBA = 180° [By angle sum property of triangle]

⇒ 90° + 34° + ∠CBA = 180°

⇒ ∠CBA = 180° - 124° = 56°.

Hence, ∠CBA = 56°.

(ii) From figure,

∠QCB = ∠CAB = 34° [Angle in alternate segment are equal.]

⇒ ∠CBQ + ∠CBA = 180° [Linear pairs]

⇒ ∠CBQ + 56° = 180°

⇒ ∠CBQ = 124°.

In △CBQ,

⇒ ∠CBQ + ∠QCB + ∠CQB = 180° [By angle sum property of triangle]

⇒ 124° + 34° + ∠CQB = 180°

⇒ ∠CQB = 180° - 158° = 22°.

Hence, ∠CQB = 22°.

In the given figure, O is the center of the circle. The tangents at B and D intersect each other at point P. If AB is parallel to CD and ∠ABC = 55°, find :

(i) ∠BOD

(ii) ∠BPD

Answer

(i) From figure,

∠BCD = ∠ABC = 55° [Alternate angles are equal.]

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠BOD = 2∠BCD = 2 x 55° = 110°.

Hence, ∠BOD = 110°.

(ii) We know that,

A tangent line is always at a right angle to the radius of the circle at the point of tangency.

∴ ∠OBP = 90° and ∠ODP = 90°.

In quadrilateral ODPB,

⇒ ∠BOD + ∠OBP + ∠ODP + ∠BPD = 360° [Angle sum property of quadrilateral]

⇒ 110° + 90° + 90° + ∠BPD = 360°

⇒ ∠BPD = 360° - 290° = 70°.

Hence, ∠BPD = 70°.

In the following figure, PQ = QR, ∠RQP = 68°, PC and CQ are tangents to the circle with center O.

Calculate the values of :

(i) ∠QOP

(ii) ∠QCP

Answer

(i) Given,

PQ = QR

∴ ∠PRQ = ∠QPR [Angles opposite to equal sides are equal in a triangle.]

In △PQR,

⇒ ∠PRQ + ∠QPR + ∠RQP = 180°

⇒ ∠PRQ + ∠PRQ + 68° = 180°

⇒ 2∠PRQ = 180° - 68°

⇒ 2∠PRQ = 112°

⇒ ∠PRQ = $\dfrac{112°}{2}$

⇒ ∠PRQ = 56°.

We know that,

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

∴ ∠QOP = 2∠PRQ = 2 x 56 = 112°.

Hence, ∠QOP = 112°.

(ii) We know that,

A tangent line is always at a right angle to the radius of the circle at the point of tangency.

∴ ∠OPC = 90° and ∠OQC = 90°.

In quadrilateral OQCP,

⇒ ∠QOP + ∠OPC + ∠OQC + ∠QCP = 360° [Angle sum property of quadrilateral]

⇒ 112° + 90° + 90° + ∠QCP = 360°

⇒ ∠QCP = 360° - 292° = 68°.

Hence, ∠QCP = 68°.

In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively. Given that AB = 8 cm, calculate PQ.

Answer

Since, AC is a tangent to the circle with center P at point A.

∴ ∠PAB = 90°.

Since, AC is a tangent to the circle with center Q at point C.

∴ ∠QCB = 90°.

In △PAB and △QCB,

⇒ ∠PAB = ∠QCB (Both equal to 90°)

⇒ ∠PBA = ∠QBC (Vertically opposite angles are equal)

⇒ △PAB ~ △QCB.

In right angle △PAB,

$\phantom{\Rightarrow} \text{PB}^2 = \text{PA}^2 + \text{AB}^2 \\[1em] \Rightarrow \text{PB} = \sqrt{\text{PA}^2 + \text{AB}^2} \\[1em] \Rightarrow \text{PB} = \sqrt{6^2 + 8^2} \\[1em] \Rightarrow \text{PB} = \sqrt{36 + 64} \\[1em] \Rightarrow \text{PB} = \sqrt{100} \\[1em] \Rightarrow \text{PB} = 10 \text{ cm}$

We know that,

In similar triangles ratio of corresponding sides are equal.

$\Rightarrow \dfrac{PA}{QC} = \dfrac{PB}{QB} \\[1em] \Rightarrow \dfrac{6}{3} = \dfrac{10}{QB} \\[1em] \Rightarrow QB = \dfrac{3 \times 10}{6} \\[1em] \Rightarrow QB = \dfrac{30}{6} \\[1em] \Rightarrow QB = 5 \text{ cm}.$

From figure,

QP = QB + PB = 5 + 10 = 15 cm.

Hence, QP = 15 cm.

In the figure, given below, O is the center of the circumcircle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.

Answer

YT and XT are tangents to the circle.

∴ ∠OYT = 90° and ∠OXT = 90°.

In quadrilateral OYTX,

⇒ ∠XOY + ∠OYT + ∠OXT + ∠XTY = 360°

⇒ ∠XOY + 90° + 90° + 80° = 360°

⇒ ∠XOY = 360° - 260° = 100°.

From figure,

⇒ ∠XOZ + ∠YOZ + ∠XOY = 360°

⇒ 140° + ∠YOZ + 100° = 360°

⇒ ∠YOZ = 360° - 240° = 120°.

We know that,

When two angles are subtended by the same arc, the angle at the centre of a circle is twice the angle at the circumference.

∴ ∠YOZ = 2∠ZXY

⇒ ∠ZXY = $\dfrac{1}{2}$∠YOZ = $\dfrac{1}{2} \times 120°$ = 60°.

Hence, ∠ZXY = 60°.

In the given circle with centre O, angle ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.

Answer

From figure,

⇒ ∠ADC + ∠ABC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠ADC + 100° = 180°

⇒ ∠ADC = 180° - 100°

⇒ ∠ADC = 80°.

In △ADC,

⇒ ∠ADC + ∠CAD + ∠ACD = 180° [By angle sum property of triangle]

⇒ 80° + ∠CAD + 40° = 180°

⇒ ∠CAD = 180° - 120° = 60°.

From figure,

⇒ ∠DCT = ∠CAD = 60° [Angles in alternate segment are equal].

Hence, ∠DCT = 60° and ∠ADC = 80°.

In the figure given below, O is the center of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.

Answer

As, SP is tangent to the circle.

∴ ∠TSR = 90°.

In △TSR,

⇒ ∠TSR + ∠STR + ∠SRT = 180° [By angle sum property of triangle]

⇒ 90° + x + 65° = 180°

⇒ x = 180° - 155° = 25°.

We know that,

Angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.

∴ ∠SOQ = 2∠STQ

⇒ y = 2x = 2(25°) = 50°.

In △OSP,

⇒ ∠OSP + ∠SOP + ∠SPO = 180° [By angle sum property of triangle]

⇒ 90° + y + z = 180°

⇒ 90° + 50° + z = 180°

⇒ z = 180° - 140° = 40°.

Hence, x = 25°, y = 50° and z = 40°.