In a cyclic quadrilateral, the sum of opposite angles is 180°.
In quadrilateral ABCD,
⇒ ∠ABC + ∠ADC = 180°
⇒ ∠x + ∠y = 180°
⇒ 80° + ∠y = 180°
⇒ ∠y = 100°.
Substitute values in L.H.S of equation 3∠y − 2∠x = 140° :
= 3(100°) − 2(80°)
= 300° − 160°
= 140°.
As, L.H.S = R.H.S
Hence, proved that 3∠y − 2∠x = 140°.
10(ii)Exercise 18
In the given figure, O is the centre of the circle. If ∠CBD = 25° and ∠APB = 120°, find ∠ADB.
Show SolutionHide Solution↓
From figure,
⇒ ∠CPB + ∠APB = 180° [Linear pair]
⇒ ∠CPB = 180° - ∠APB
⇒ ∠CPB = (180° − 120°) = 60°.
By angle sum property of a triangle we get,
⇒ ∠PCB + ∠CPB + ∠PBC = 180°
⇒ ∠PCB + 60° + 25° = 180°
⇒ ∠PCB + 85° = 180°
⇒ ∠PCB = 180° - 85°
⇒ ∠PCB = 95°
⇒ ∠ADB = ∠ACB = 95° [Angles in same segment of a circle are equal]
Hence, the value of ∠ADB = 95°.
11Exercise 18
In the given figure, O is the centre of the circle, ∠BAD = 75° and chord BC = chord CD. Find : (i) ∠BOC (ii) ∠OBD (iii) ∠BCD
Show SolutionHide Solution↓
(i) As equal chords of a circle subtend equal angles at the center and chord BC = chord CD, so ∠BOC = ∠COD.
∠BOD = 2 × ∠BAD
∠BOD = 2 × 75°
∠BOD = 150°
∠BOC = $\frac{1}{2}$ ∠BOD
= $\frac{1}{2}$ × 150° = 75°.
Hence, the value of ∠BOC = 75°.
(ii) Join BD.
Since, OB = OD
∴ ∠OBD = ∠ODB = x
Since sum of angles of triangle = 180°
In △OBD
⇒ ∠BOD + ∠OBD + ∠ODB = 180°
⇒ 150° + x + x = 180°
⇒ 150° + 2x = 180°
⇒ 2x = 180° - 150°
⇒ 2x = 30°
⇒ x = 15°.
Hence, the value of ∠OBD = 15°.
(iii) ABCD is a cyclic quadrilateral as all of its vertices lie on the circumference of the circle.
We know that sum of opposite angles of a cyclic quadrilateral = 180°.
⇒ ∠BCD + ∠BAD = 180°
⇒ ∠BCD + 75° = 180°
⇒ ∠BCD = 180° - 75°
⇒ ∠BCD = 105°.
Hence, the value of ∠BCD = 105°.
12Exercise 18
In the figure given, O is the centre of the circle. ∠DAE = 70°, find giving suitable reasons, the measure of :
(i) ∠BCD
(ii) ∠BOD
(iii) ∠OBD
Show SolutionHide Solution↓
(i) Given,
∠DAE = 70°
From figure,
⇒ ∠DAE + ∠BAD = 180° [Linear pairs]
⇒ 70° + ∠BAD = 180°
⇒ ∠BAD = 180° - 70° = 110°.
We know that,
Sum of opposite angles in a cyclic quadrilateral = 180°
⇒ ∠BCD + ∠BAD = 180°
⇒ ∠BCD + 110° = 180°
⇒ ∠BCD = 180° - 110° = 70°.
Hence, ∠BCD = 70°.
(ii) We know that,
Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.
⇒ ∠BOD = 2∠BCD = 2 × 70° = 140°.
Hence, ∠BOD = 140°.
(iii) In △OBD,
OB = OD [Radius of same circle]
∠OBD = ∠ODB = x.
⇒ ∠OBD + ∠ODB + ∠BOD = 180°
⇒ x + x + 140° = 180°
⇒ 2x = 180° - 140°
⇒ 2x = 40°
⇒ x = $\frac{40^{\circ}}{2}$ = 20°.
∴ ∠OBD = 20°.
Hence, ∠OBD = 20°.
13Exercise 18
In the given figure, AOB is a diameter of the circle with centre O and ∠AOC = 100°, find ∠BDC.
Show SolutionHide Solution↓
From figure,
∠AOC + ∠BOC = 180° [Linear pair]
100° + ∠BOC = 180°
∠BOC = 180° - 100°
∠BOC = 80°.
We know that,
Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.
⇒ ∠BOC = 2∠BDC
⇒ 80° = 2∠BDC
⇒ ∠BDC = $\frac{80^{\circ}}{2}$ = 40°
Hence, ∠BDC = 40°.
14Exercise 18
In the given figure, find whether the points A, B, C, D are concyclic when:
(i) x = 70
(ii) x = 80
Show SolutionHide Solution↓
From the figure,
∠ABC + ∠CBE = 180° [Linear pair]
∠ABC + 110° = 180°
∠ABC = 70°
For the points A, B, C, D to be concyclic, ∠ABC + ∠ADC must be equal to 180°.
(i) When x = 70
x° + ∠ADC = 180° [Linear pair]
∠ADC = 180° − x° = 180° - 70° = 110°.
∠ABC + ∠ADC = 70° + 110° = 180°
Since the sum of opposite angles is 180°, the points A, B, C, D are concyclic when x = 70.
Hence, yes the points A, B, C, D are concyclic when x = 70.
(ii) When x = 80
x° + ∠ADC = 180° [Linear pair]
∠ADC = 180° − 80° = 100°
Check for con-cyclicity,
∠ABC + ∠ADC = 70° + 100° = 170°
Since the sum of opposite angles is not 180°, the points A, B, C, D are not concyclic when x = 80.
Hence, no the points A, B, C, D are not concyclic when x = 80.
15Exercise 18
In the adjoining figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find: (i) ∠BCD (ii) ∠ADB Hence, show that AC is a diameter.
Show SolutionHide Solution↓
(i) We know that,
Sum of opposite angles in a cyclic quadrilateral = 180°.
In cyclic quadrilateral ABCD,
∴ ∠BCD + ∠BAD = 180°
⇒ ∠BCD + 65° = 180°
⇒ ∠BCD = 180° - 65° = 115°.
Hence, ∠BCD = 115°.
(ii) In △ABD,
⇒ ∠ADB + ∠BAD + ∠DBA = 180° [Angle sum property of triangle]
⇒ ∠ADB + 65° + 70° = 180°
⇒ ∠ADB + 135° = 180°
⇒ ∠ADB = 180° - 135° = 45°.
We know that,
Angles in same segment are equal.
∴ ∠ACB = ∠ADB = 45°.
From figure,
∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°.
Since, angle in a semi-circle is a right angle. Thus, AC is the diameter.
Hence, ∠ADB = 45° and proved that AC is a diameter.
16Exercise 18
In the given figure, ABCD is a cyclic quadrilateral in which ∠CAD = 25°, ∠ADB = 35° and ∠ABD = 50°. Calculate:
(i) ∠CBD
(ii) ∠CAB
(iii) ∠ACB
Show SolutionHide Solution↓
(i) We know that,
Angles in same segment are equal.
∠CBD = ∠CAD = 25°
Hence, ∠CBD = 25°.
(ii) We know that,
Angles in same segment are equal.Therefore,
⇒ ∠ACD = ∠ABD = 50°
⇒ ∠ACB = ∠ADB = 35°
From figure
⇒ ∠BCD = ∠ACD + ∠ACB
⇒ ∠BCD = 35° + 50°
⇒ ∠BCD = 85°.
We know that,
Sum of opposite angles in a cyclic quadrilateral = 180°.
⇒ ∠DAB + ∠DCB = 180°
⇒ ∠DAB = 180° - 85°
⇒ ∠DAB = 95°.
From figure,
⇒ ∠CAB = ∠DAB - ∠DAC
= 95° - 25°
= 70°.
Hence, ∠CAB = 70°.
(iii) Angles in same segment are equal. Therefore,
∠ACB = ∠ADB = 35°
Hence, ∠ACB = 35°.
17Exercise 18
In the figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find :
(i) ∠CAD
(ii) ∠CBD
(iii) ∠ADC
Show SolutionHide Solution↓
(i) We know that,
Exterior angle of a cyclic quadrilateral is equal to interior opposite angle.
∠BAD = Exterior ∠BCE = 80°.
From figure,
∠CAD = ∠BAD - ∠BAC = 80° - 25° = 55°.
Hence, ∠CAD = 55°.
(ii) We know that,
Angles in same segment are equal.
∴ ∠CBD = ∠CAD = 55°.
Hence, ∠CBD = 55°.
(iii) Since, AB ∥ DC,
∠ACD = ∠BAC = 25° [Alternate angles are equal]
In triangle ADC,
By angle sum property of triangle,
⇒ ∠ADC + ∠CAD + ∠ACD = 180°
⇒ ∠ADC + 55° + 25° = 180°
⇒ ∠ADC + 80° = 180°
⇒ ∠ADC = 180° - 80°
⇒ ∠ADC = 100°.
Hence, ∠ADC = 100°.
18Exercise 18
In the adjoining figure of a circle with centre O and diameter AD, ∠BED = 70° and BC is parallel to AD. Find:
(i) ∠BAD
(ii) ∠BOD
(iii) ∠DBC
(iv) ∠DCF
Show SolutionHide Solution↓
(i) Given,
∠BED = 70°
We know that,
Angles in the same segment of a circle are equal.
∠BAD = ∠BED = 70°.
Hence, ∠BAD = 70°.
(ii) We know that,
The angle which an arc of a circle subtends at the center is double which it subtends at any point on the remaining part of the circumference.
Therefore,
∠BOD = 2∠BAD = 2 × 70° = 140°.
Hence, ∠BOD = 140°.
(iii) We know that,
Angle in a semi-circle is a right angle triangle.
∠ABD = 90°
In △ABD,
⇒ ∠BAD + ∠BDA + ∠ABD = 180°
⇒ 70° + ∠BDA + 90° = 180°
⇒ 160° + ∠BDA = 180°
⇒ ∠BDA = 180° - 160°
⇒ ∠BDA = 20°
From figure,
BC || AD
∴ ∠DBC = ∠BDA = 20° (Alternate angles are equal).
Hence, ∠DBC = 20°.
(iv) The figure ABCD is a cyclic quadrilateral
We know that,
The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
∠DCF = ∠BAD = 70°.
Hence, ∠DCF = 70°.
19Exercise 18
In the given figure, AB is a diameter of a circle with centre O and chord ED is parallel to AB and ∠EAB = 65°. Calculate : (i) ∠EBA (ii) ∠BED (iii) ∠BCD
Show SolutionHide Solution↓
(i) We know that,
Angle in semi-circle is a right angle.
∴ ∠AEB = 90°.
In △AEB,
⇒ ∠AEB + ∠EBA + ∠EAB = 180°
⇒ 90° + ∠EBA + 65° = 180°
⇒ 155° + ∠EBA = 180°
⇒ ∠EBA = 180° - 155° = 25°.
Hence, ∠EBA = 25°.
(ii) ∠BED = ∠EBA = 25° Alternate interior angles ED ∥ AB, EB as a transversal
Hence, ∠BED = 25°.
(iii) As, AB ∥ ED
∴ ∠DEB = ∠EBA = 25° [Alternate angles]
BCDE is a cyclic quadrilateral.
∴ ∠DEB + ∠BCD = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°.]
⇒ 25° + ∠BCD = 180°
⇒ ∠BCD = 180° - 25° = 155°.
Hence, ∠BCD = 155°.
20Exercise 18
In the given figure, O is the centre of a circle and ABE is a straight line. If ∠CBE = 55°, find :
(i) ∠ADC
(ii) ∠ABC
(iii) the value of x.
Show SolutionHide Solution↓
(i) We know that,
In a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
∠ADC = ∠CBE = 55°
Hence, ∠ADC = 55°.
(ii) From figure,
∠CBE + ∠ABC = 180° [Linear pairs]
∠ABC = 180° - 55°
∠ABC = 125°.
Hence, ∠ABC = 125°.
(iii) We know that,
Angle which an arc subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.
∠AOC = 2∠ADC
∠AOC = 2(55°)
∠AOC = 110°
The reflex angle AOC,
x° = 360° - angle AOC
x° = 360° - 110°
x° = 250°.
Hence, x = 250.
21Exercise 18
In the given figure AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that Δ AEB is isosceles.
Show SolutionHide Solution↓
We know that,
Exterior angle of a cyclic quadrilateral is equal to interior opposite angle.
∠EDC = ∠A ....(i)
∠DCE = ∠B ....(ii)
AB ∥ CD
∠EDC = ∠B [Corresponding angles] ......(iii)
∠DCE = ∠A [Corresponding angles] .......(iv)
From (i), (ii), (iii) and (iv) we get :
∴ ∠A = ∠B
BE = AE [Sides opposite to equal angles are equal]
Hence, proved that ΔAEB is isosceles.
22Exercise 18
In the given figure, the two circles intersect at P and Q. If ∠A = 80° and ∠D = 84°, calculate :
(i) ∠QBC
(ii) ∠BCP
Show SolutionHide Solution↓
PQAD is a cyclic quadrilateral as all vertices lie on the circumference of the circle.
Sum of opposite angles of cyclic quadrilateral = 180°
⇒ ∠DAQ + ∠DPQ = 180°
⇒ 80° + ∠DPQ = 180°
⇒ ∠DPQ = 180° - 80°
⇒ ∠DPQ = 100°.
Also,
⇒ ∠PDA + ∠PQA = 180°
⇒ 84° + ∠PQA = 180°
⇒ ∠PQA = 180° - 84°
⇒ ∠PQA = 96°.
Since exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.
(i) ∠QBC = ∠DPQ = 100°.
Hence, ∠QBC = 100°.
(ii) ∠BCP = ∠PQA = 96°.
Hence, ∠BCP = 96°.
23Exercise 18
In the given figure, O is the centre of the circle. If ∠AOD = 140° and ∠CAB = 50°, calculate :
(i) ∠EDB
(ii) ∠EBD
Show SolutionHide Solution↓
(i) We know that,
An Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle. ∠EDB is the exterior angle at vertex D. The interior opposite angle is ∠CAB.
⇒ ∠EBD = ∠CAB = 50°
Hence, the value of ∠EDB = 50°.
(ii) We know that,
⇒ OD = OB [radii of same circle]
From figure,
⇒ ∠AOD + ∠DOB = 180° [Linear pairs]
⇒ 140° + ∠DOB = 180°
⇒ ∠DOB = 180° - 140°
⇒ ∠DOB = 40°
In ΔODB,
By angle sum property of triangle,
⇒ ∠DOB + ∠ODB + ∠OBD = 180°
⇒ 40° + 2∠OBD = 180°
⇒ 2∠OBD = 180° - 40°
⇒ 2∠OBD = 140°
⇒ ∠OBD = $\frac{180^{\circ}}{2}$
⇒ ∠OBD = 70°
From figure,
⇒ ∠OBD + ∠EBD = 180° [Linear pairs]
⇒ 70° + ∠EBD = 180°
⇒ ∠EBD = 180° - 70°
⇒ ∠EBD = 110°
Hence, the value of ∠EBD = 110°.
24Exercise 18
In the given figure, AB is a diameter of a circle with centre O. If ADF and CBF are straight lines, meeting at F such that ∠BAD = 35° and ∠BFD = 25°, find :
(i) ∠DCB
(ii) ∠DBC
(iii) ∠BDC
Show SolutionHide Solution↓
(i) We know that,
Angles in same segment are equal.
∠DCB = ∠BAD = 35°
Hence, the value of ∠DCB = 35°.
(ii) From figure,
∠ADB = 90° (∵ Angle in semicircle is 90°.)
∴ ∠BDF = 90°
In △DBF,
⇒ ∠DBF + ∠DFB + ∠BDF = 180° (∵ By angle sum triangle property).
⇒ ∠DBF = 180° - (∠DFB + ∠BDF)
⇒ ∠DBF = 180° - (90° + 25°)
⇒ ∠DBF = 65°.
⇒ ∠DBF + ∠DBC = 180° (Linear pairs)
⇒ ∠DBC = 180° - 65°
⇒ ∠DBC = 115°.
Hence, the value of ∠DBC = 115°.
(iii) In △DCB,
⇒ ∠DCB + ∠CBD + ∠BDC = 180° (∵ By angle sum triangle property).
⇒ ∠BDC = 180° - (∠DCB + ∠CBD)
⇒ ∠BDC = 180° - (35° + 115°)
⇒ ∠BDC = 180° - (35° + 115°)
⇒ ∠BDC = 30°.
Hence, the value of ∠BDC = 30°.
25Exercise 18
In the given figure, the straight lines AB and CD pass through the centre O of the circle. If ∠AOD = 75° and ∠OCE = 40°, find :
(i) ∠CDE
(ii) ∠OBE.
Show SolutionHide Solution↓
(i) Since, AB and CD pass through the center, thus they are the diameters of circle.
In △CED,
∠CED = 90° (∵ angle in semicircle is 90°.)
We know that sum of angles of a triangle is 180°.
⇒ ∠CED + ∠DCE + ∠CDE = 180°.
⇒ 90° + 40° + ∠CDE = 180°
⇒ ∠CDE + 130° = 180°
⇒ ∠CDE = 180° - 130°
⇒ ∠CDE = 50°.
Hence, ∠CDE = 50°.
(ii) From figure,
⇒ ∠AOD + ∠DOB = 180° (Linear pairs)
⇒ 75° + ∠DOB = 180°
⇒ ∠DOB = 180° - 75°
⇒ ∠DOB = 105°.
In △DOB,
∠ODB = ∠CDE = 50°
We know that,
Sum of angles of a triangle is 180°.
⇒ ∠DOB + ∠ODB + ∠DBO = 180°.
⇒ 105° + 50° + ∠DBO = 180°
⇒ ∠DBO + 155° = 180°
⇒ ∠DBO = 180° - 155°
⇒ ∠DBO = 25°.
From figure,
∠OBE = ∠DBO
∴ ∠OBE = 25°.
Hence, ∠OBE = 25°.
26Exercise 18
In the adjoining figure, AB = AC = CD and ∠ADC = 35°. Calculate :
(i) ∠ABC
(ii) ∠BEC
Show SolutionHide Solution↓
(i) Given,
AC = CD
In triangle ACD,
∠DAC = ∠ADC = 35° [Angles opposite to equal sides in a triangle are equal]
In △ACD,
⇒ ∠DAC + ∠ADC + ∠ACD = 180°
⇒ 35° + 35° + ∠ACD = 180°
⇒ 70° + ∠ACD = 180°
⇒ ∠ACD = 180° - 70° = 110°.
From figure,
⇒ ∠ACB + ∠ACD = 180° [Linear pair]
⇒ ∠ACB + 110° = 180°
⇒ ∠ACB = 180° - 110° = 70°.
Given,
AB = AC
∴ ∠ABC = ∠ACB = 70°. [As angles opposite to equal sides are equal]
Hence, ∠ABC = 70°.
(ii) In △ABC,
⇒ ∠BAC + ∠ACB + ∠ABC = 180° [Angle sum property of triangle]
⇒ ∠BAC + 70° + 70° = 180°
⇒ ∠BAC + 140° = 180°
⇒ ∠BAC = 180° - 140° = 40°.
We know that,
Angles in same segment are equal.
⇒ ∠BEC = ∠BAC = 40°.
Hence, ∠BEC = 40°.
27Exercise 18
The exterior angles B and C in ΔABC are bisected to meet at a point P. Prove that ∠BPC = 90° − $\frac{A}{2}$. Is ABPC a cyclic quadrilateral ?
Show SolutionHide Solution↓
We know that,
An exterior angle of a triangle equals to the sum of the two interior opposite angles.
∠CBD = ∠A + ∠C
∠BCE = ∠A + ∠B.
Since BP bisects the exterior angle ∠CBD :
∠PBC = $\frac{1}{2}$ ∠CBD
Similarly, since CP bisects the exterior angle ∠BCE :
Opposite angles of a cyclic quadrilateral ADBC are supplementary.
∠ACB + ∠ADB = 180°
∠ADB = 180° - 30°
∠ADB = 150°.
Thus, angle subtended by this chord at the minor arc of the circle is 150°.
Hence, option 4 is the correct option.
301 Mark · Multiple Choice Questions
ABCD is a cyclic quadrilateral such that ∠ADB = 45° and ∠DCA = 55°, then ∠DAB is equal to :
(a)55°
(b)45°
(c)100°
(d)80°
Show SolutionHide Solution↓
∠ACB = ∠ADB = 45° [Angles in the same segment]
From figure,
∠BCD = ∠ACD + ∠ACB
∠BCD = 55° + 45° = 100°
In a cyclic quadrilateral, the sum of opposite angles is 180°.
∠DAB + ∠BCD = 180°
∠DAB = 180° - ∠BCD
∠DAB = 180° - 100° = 80°.
Hence, option 4 is the correct option.
311 Mark · Multiple Choice Questions
In the given figure, if ∠ACD = 30° and ∠BPD = 100°, then ∠CDB is equal to :
(a)30°
(b)40°
(c)50°
(d)60°
Show SolutionHide Solution↓
∠ABD = ∠ACD = 30° [Angles in the same segment]
From figure,
∠PBD = ∠ABD = 30°
In ΔPBD,
By angle sum property,
∠PBD + ∠BPD + ∠PDB = 180°
30° + 100° + ∠PDB = 180°
∠PDB = 180° - 130°
∠PDB = 50°.
From figure,
∠CDB = ∠PDB = 50°.
Hence, option 3 is the correct option.
321 Mark · Multiple Choice Questions
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 125°, then ∠BAC is equal to :
(a)35°
(b)40°
(c)45°
(d)55°
Show SolutionHide Solution↓
Opposite angles of a cyclic quadrilateral are supplementary.
So,
∠ABC + ∠ADC = 180°
∠ABC = 180° − 125° = 55°.
∠ACB = 90° [Angle in semicircle is a right angle]
In △ABC,
By angle sum property of triangle,
∠BAC + ∠ABC + ∠ACB = 180°
∠BAC + 55° + 90° = 180°
∠BAC = 180° − 145° = 35°.
Hence, option 1 is the correct option.
331 Mark · Multiple Choice Questions
In the given figure, if O is the centre of the circle, then the value of x is :
(a)15°
(b)18°
(c)20°
(d)24°
Show SolutionHide Solution↓
From figure,
∠DAB = 90° [Angle in the semicircle is a right angle]
In the figure, both ∠ADB and ∠ACB are subtended by the same arc AB. Therefore:
∠ADB = ∠ACB = 2x
In triangle ADB,
∠DAB + ∠ADB + ∠ABD = 180°
90° + 2x + 3x = 180°
5x = 90°
x = 18°.
Hence, option 2 is the correct option.
341 Mark · Multiple Choice Questions
In the given figure, O is the centre of the circle. If the length of chord AB is equal to the radius of the circle; then ∠ACB is equal to :
(a)30°
(b)45°
(c)60°
(d)40°
Show SolutionHide Solution↓
If the length of chord AB is equal to the radius of the circle; then △AOB is equilateral triangle.
∠AOB = 60°
We know that,
The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.
∴ ∠ACB = $\frac{1}{2}$ ∠AOB
∠ACB = $\frac{60^{\circ}}{2}$ = 30°.
Hence, option 1 is the correct option.
351 Mark · Multiple Choice Questions
The value of x in the following figure is :
(a)80°
(b)57°
(c)87°
(d)96°
Show SolutionHide Solution↓
Arc AD subtends ∠ABD and ∠ACD. Therefore,
∠ABD = ∠ACD = 55° [From Figure]
Arc AB subtends ∠BCA and ∠BDA. Therefore,
∠BCA = ∠BDA = 32° [From Figure]
From figure,
∠BCD = ∠BCA + ∠ACD
∠BCD = 32° + 55°
∠BCD = 87°.
Hence, option 3 is the correct option.
361 Mark · Multiple Choice Questions
In the given figure, if O is the centre of the circle, then the value of x is :
(a)40°
(b)50°
(c)70°
(d)30°
Show SolutionHide Solution↓
Join AC and BD.
∠BDC = 90° (Angle in semicircle is right angle)
OA = OB (Radii of same circle)
∠OBA = ∠OAB = 50° (Angles opposite to equal sides are equal in a triangle)
In triangle OAB,
∠AOB + ∠OAB + ∠OBA = 180°
∠AOB = 180° - (∠OAB + ∠OBA)
∠AOB = 180° - (50° + 50°)
∠AOB = 80°.
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.
∠ACB = $\frac{1}{2}$ ∠AOB
∠ACB = $\frac{80°}{2}$
∠ACB = 40°
∠ADB = ∠ACB = 40° [Angle in same segment are equal]
From figure,
∠BDC = ∠ADB + x
90° = 40° + x
x = 50°.
Hence, option 2 is the correct option.
371 Mark · Multiple Choice Questions
In the given figure, O is centre of the circle. If ∠AOC = 130°, then the value of x is :
(a)25°
(b)50°
(c)65°
(d)40°
Show SolutionHide Solution↓
Join AD.
∠AOC = 130°
∠ADB = 90°
We know that,
The angle which an arc subtends at the center is double that which it subtends at any point on the remaining part of the circumference.
∴ ∠ADC = $\frac{1}{2}$ ∠AOC
∠ADC = $\frac{130^{\circ}}{2}$ = 65°
At point D,
∠ADB = ∠ADC + ∠CDB
90° = 65° + x
x = 25°
Hence, option 1 is the correct option.
381 Mark · Multiple Choice Questions
In the given figure, O is the centre of the circle, ∠AOB = 40° and ∠BDC = 100°. The measure of ∠OBC is :
(a)40°
(b)80°
(c)60°
(d)20°
Show SolutionHide Solution↓
We know that,
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠ACB = $\frac{1}{2}$ ∠AOB
∠ACB = $\frac{40^{\circ}}{2}$ = 20°
In ΔBDC, the sum of all angles is 180°
∠OBC + ∠BDC + ∠ACB = 180°
∠OBC + 100° + 20° = 180°
∠OBC = 180° - 120°
∠OBC = 60°
Hence, option 3 is the correct option.
391 Mark · Multiple Choice Questions
ABCD is a cyclic quadrilateral. If ∠BAD = (2x + 5)° and ∠BCD = (x + 10)°, then x is equal to :
(a)65°
(b)45°
(c)55°
(d)5°
Show SolutionHide Solution↓
In a cyclic quadrilateral, opposite angles are supplementary.
Since ∠BAD and ∠BCD are opposite angles,
(2x + 5)° + (x + 10)° = 180°
3x + 15 = 180°
3x = 165°
x = 55°
Hence, option 3 is the correct option.
1Assertion-Reason Type Questions
Assertion (A): In the given figure, if O is the centre of the circle, then ∠ACB = 40°.
Reason (R): Angle at the centre is double the angle at the remaining part of the circle.
(a)Both A and R are true, and R is the correct explanation of A.
(b)Both A and R are true, but R is not the correct explanation of A.
(c)A is true, but R is false.
(d)A is false, but R is true.
Show SolutionHide Solution↓
From figure,
OA = OB (Radii of same circle)
∠OBA = ∠OAB = 50° (As angles opposite to equal sides in a triangle are equal)
∠OBA + ∠OAB + ∠AOB = 180° [Angle sum property of a triangle]
50° + 50° + ∠AOB = 180°
100° + ∠AOB = 180°
∠AOB = 180° - 100°
∠AOB = 80°.
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.
So, reason (R) is true.
∠AOB = 2∠ACB
∠ACB = $\frac{80°}{2}$ = 40°.
So, Assertion (A) is true.
Both A and R are true, and R correctly explains A.
Hence, option 1 is the correct option.
2Assertion-Reason Type Questions
Assertion (A): In the figure, ∠ACB = 70°.
Reason (R): Opposite angles of a cyclic quadrilateral are equal.
(a)Both A and R are true, and R is the correct explanation of A.
(b)Both A and R are true, but R is not the correct explanation of A.
(c)A is true, but R is false.
(d)A is false, but R is true.
Show SolutionHide Solution↓
In triangle ABD,
∠DAB = 60°
∠DBA = 50°
From figure,
⇒ ∠DAB + ∠DBA + ∠ADB = 180° [Angle sum property of a triangle]
⇒ 60° + 50° + ∠ADB = 180°
⇒ 110° + ∠ADB = 180°
⇒ ∠ADB = 180° - 110°
⇒ ∠ADB = 70°
Angle in same segment are equal.
∠ADB = ∠ACB = 70°
So, assertion (A) is true.
We know that,
In cyclic quadrilateral opposite angles are supplementary (sum = 180°), not equal.
So, reason (R) is false.
A is true, R is false.
Hence, option 3 is the correct option.
3Assertion-Reason Type Questions
Assertion (A): In the figure, if AB is a diameter of the circle, then ∠BAC = 40°.
Reason (R): Angles in the same segment of a circle are equal.
(a)Both A and R are true, and R is the correct explanation of A.
(b)Both A and R are true, but R is not the correct explanation of A.
(c)A is true, but R is false.
(d)A is false, but R is true.
Show SolutionHide Solution↓
In △ABC,
∠ACB = 90° [Angle in a semicircle]
Since ABCD is a cyclic quadrilateral, the sum of its opposite angles is 180°.
⇒ ∠ADC + ∠ABC = 180°
⇒ 130° + ∠ABC = 180°
⇒ ∠ABC = 50°
In △ABC,
⇒ ∠BAC + ∠ABC + ∠ACB = 180° [Angle sum property of a triangle]
⇒ ∠BAC = 180° - (50° + 90°)
⇒ ∠BAC = 40°.
So, Assertion (A) is true.
Angles in the same segment of a circle are equal, so Reason (R) is a true statement. However, ∠BAC = 40° is obtained using the angle in a semicircle together with the cyclic-quadrilateral property, and not by the equality of angles in the same segment. So R is not the correct explanation of A.
Both A and R are true, but R is not the correct explanation of A.
Hence, option 2 is the correct option.
4Assertion-Reason Type Questions
Assertion (A): In the figure, if O is the centre of the circle, then ∠BCD = 80°.
Reason (R): Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
(a)Both A and R are true, and R is the correct explanation of A.
(b)Both A and R are true, but R is not the correct explanation of A.
(c)A is true, but R is false.
(d)A is false, but R is true.
Show SolutionHide Solution↓
From figure,
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Reflex AOB = 360° - 100° = 260°
∠ACB = $\frac{1}{2}$ Reflex ∠AOB
∠ACB = $\frac{260 ^{\circ}}{2}$
∠ACB = 130°
∠ACB + ∠BCD = 180° [Linear pair]
∠BCD = 180° - 130°
∠BCD = 50°
So assertion (A) is false.
We know that,
In, cyclic quadrilaterals if one side is extended, the exterior angle formed is equal to the interior opposite angle.
So, reason (R) is true.
A is false, R is true
Hence, option 4 is the correct option.
1Analytical and Application Based Questions
AB and CD intersect at the center O of the circle given in the above diagram. If ∠EBA = 33° and ∠EAC = 82°, find
(a) ∠BAE
(b) ∠BOC
(c) ∠ODB
Show SolutionHide Solution↓
(a) In △ BAE,
⇒ ∠AEB = 90° (Angle in a semicircle is a right angle)
By angle sum property of triangle,
⇒ ∠AEB + ∠EBA + ∠BAE = 180°
⇒ 90° + 33° + ∠BAE = 180°
⇒ ∠BAE + 123° = 180°
⇒ ∠BAE = 180° - 123° = 57°.
Hence, ∠BAE = 57°.
(b) From figure,
⇒ ∠EAC = ∠EAB + ∠BAC
⇒ 82° = 57° + ∠BAC
⇒ ∠BAC = 82° - 57° = 25°.
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.
We use essential cookies for sign-in and cookieless traffic measurement. Optional analytics help us improve the question bank; nothing optional loads until you allow it.
