Question 1:In the given figure ,PQ is a diameter of a circle with centre O . If âˆ PRQ = 65Â°,âˆ SPR = 40Â° and âˆ PQM = 50Â° Â Â Â , Â Â Â Â find âˆ QPR ,âˆ QPM and âˆ PRS.

Sol.

Consider the triangle Î” PRQ.

PQ is the diameter.

The angle in a semicircle is a right angle.

â‡’ âˆ PRQ = 90Â°

By the angle sum property in Î” PRQ, we have

âˆ QPR + âˆ PRQ +âˆ PQR = 180Â°

â‡’ âˆ QPR + 90Â° + 65Â° = 180Â°

â‡’ âˆ QPR = 180Â° – 155Â° = 25Â° â€¦â€¦.(1)

Now Consider the triangle Î” PQM.

Since PQ is the diameter, âˆ PMQ = 90Â°â€™

Again by the angle sum property in Î” PRQ, we have

âˆ QPM+ âˆ PMQ +âˆ PQM = 180Â°

â‡’ âˆ QPM + 90Â° + 50Â° = 180Â°

â‡’ âˆ QPR = 180Â° – 140Â° = 40Â°

Now in quadrilateral PQRS

âˆ QPS + âˆ SRQ = 180Â°

â‡’ âˆ QPR+ âˆ RPS +âˆ PRQ + âˆ PRS = 180Â° [from (1)]

â‡’ 25Â° + 40Â° + 90Â° + âˆ PRS = 180Â°

â‡’ Â Â Â Â Â Â âˆ PRS = 180Â° – 155Â° = 25Â°

âˆ´ Â Â Â Â Â âˆ PRS = 25Â°

Question 2: In the given figure ABCD is a cyclic quadrilateral whose diagonals intersect at P such that âˆ DBC = 60Â° and âˆ BAC = 40Â° , find

- âˆ BCD
- âˆ CAD

Sol.

(i) âˆ BDC = âˆ BAC = 40Â° [angles in the same segment]

In Î” BCD, we have

âˆ BCD +âˆ BDC + âˆ DBC =180Â°

âˆ´ âˆ BCD +40Â° + 60Â° =180Â°

â‡’ âˆ BCD = 180Â° -100Â° = 80Â°

âˆ´ âˆ BCD = 80Â°

(ii) Also âˆ CAD = âˆ CBD [angles in the same segment]

âˆ CAD = 60Â° [âˆµ âˆ CBD = 60Â°]

Question 3: In the given figure, PQR is a diameter and PQRS is a cyclic quadrilateral. If âˆ PSR = 150Â°, findâˆ RPQ

Sol.

In cyclic quadrilateral PQRS

âˆ PSR + âˆ PQR = 180Â°

â‡’ 150Â° + âˆ PQR = 180Â°

â‡’ âˆ PQR = 180Â° – 150Â° = 30Â° â€¦â€¦.(1)

Also, âˆ PQR = 90Â° â€¦â€¦.(2) [angles in a semicircle]

Now in Î” PRQ, we have

âˆ PQR + âˆ PRQ + âˆ RPQ = 180Â°

â‡’ 30Â° + 90Â°+ âˆ RPQ = 180Â° [from (i) and (ii)]

â‡’ âˆ RPQ = 180Â° – 120Â° = 60Â°

âˆ´ âˆ RPQ = 60Â°

Question 4: In the given figure ABCD is a cyclic quadrilateral in which AB || DC . If Â Â Â Â Â Â âˆ BAD = 100Â° Â , find

- âˆ BCD
- âˆ ADC
- âˆ ABC

Sol.

In cyclic quadrilateral ABCD, AB || DC and âˆ BAD = 100Â°

(i) âˆ BCD + âˆ BAD = 180Â°

â‡’ âˆ BCD + 100Â° = 180Â°

â‡’ âˆ BCD = 180Â° – 100Â° = 80Â°

(ii) Also, âˆ ADC = âˆ BCD = 80Â°

âˆ´ âˆ ACD = 80Â°

(iii) âˆ ABC = âˆ BAD = 100Â°

âˆ ABC = 100Â°

Question 5: In the given figure O is the centre of the circle and arc ABC subtends an angle of 130^{0 }at the centre . If Â AB is extended to P , find âˆ PBC

Sol.

Take a point D on the major arc CA and join AD and DC

âˆ´ âˆ 2 = 2âˆ 1 [Angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.]

âˆ´ 130Â° = 2âˆ 1

â‡’ âˆ 1 = 65Â° â€¦â€¦.(1)

âˆ PBC = âˆ 1 [âˆµ Exterior angle of a cyclic quadrilateral interior opposite angle]

âˆ´ âˆ PBC = 65Â°

Question 6: In the given figure ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD and BA is produced . Ifâˆ ABC=92Â° Â and âˆ EAF = 20Â°, findâˆ BCD.

Sol.

ABCD is acyclic quadrilateral

âˆ´ âˆ ABC + âˆ ADC = 180Â°

â‡’ 92Â° + âˆ ADC = 180Â°

â‡’ âˆ ADC = 180Â° -92Â° = 88Â°

Also, AE || CD

âˆ´ âˆ EAD = âˆ ADC = 88Â°

âˆ´ âˆ BCD = âˆ DAF [âˆµ Exterior angle of a cyclic quadrilateral = interior opposite angle]

âˆ´ âˆ BCD = âˆ EAD + âˆ EAF

= 88Â° + 20Â° [âˆµâˆ FAE = 20Â° (given)]

= 108Â°

âˆ´ âˆ BCD = 108Â°

Question 7: In the given figure , BD=DC and âˆ CBD = 30Â°, find m(âˆ BAC)

Sol.

BD = DC

âˆ´ âˆ BCD = âˆ CBD = 30Â°

In Î” BCD, we have

âˆ BCD + âˆ CBD + âˆ CDB = 180Â°

â‡’ 30Â° + 30Â° + âˆ CDB = 180Â°

â‡’ âˆ CDB = 180Â° – 60Â°

= 120Â°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

âˆ CDB + âˆ BAC = 180Â°

= 180Â° -120Â° [âˆµâˆ CDB = 120Â°]

= 60Â°

âˆ´ âˆ BAC = 60Â°

Question 8: In the given figure , O is the centre of the given circle and measure of arc ABC is 100^{0}. Determine âˆ ADC Â and âˆ ABC .

Sol.

Angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.

Here arc ABC makes âˆ AOC = 100Â° at the center of the circle and âˆ ADC on the circumference of the circle.

âˆ´ âˆ AOC = 2âˆ ACD

â‡’ âˆ ACD = Â½ (âˆ AOC)

â‡’ = Â½ Ã— 100Â° [âˆ AOC = 100Â°]

â‡’ âˆ ACD = 50Â°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

âˆ ACD + âˆ ABC = 180Â°

= 180Â° – 50Â° [âˆµ âˆ ADC = 50Â°]

= 130Â°

âˆ´ âˆ ABC = 130Â°

âˆ´ âˆ ACD = 50Â° and âˆ ABC = 130Â°

Question 9: In the given figure , Î” ABC is equilateral . Find

- âˆ BDC
- âˆ BEC

Sol.

Î” ABC is an equilateral triangle.

âˆ´ Â Â Each of its angle is equal to 60Â°

â‡’ âˆ BAC = âˆ ABC = âˆ Â âˆ ACB = 60Â°

(i) Angles in the same segment of a circle are equal.

âˆ´ âˆ BDC = âˆ BAC

= 60Â° [âˆµ âˆ BAC = 60Â°]

(ii) The opposite angles of a cyclic quadrilateral are supplementary.

ABCE is a cyclic quadrilateral and thus,

âˆ BAC + âˆ BEC = 180Â°

âˆ BEC = 180Â° – 60Â° [âˆµ âˆ BAC = 60Â°]

= 120Â°

â‡’ âˆ BEC = 120Â°

Question 10: In the adjoining figure, ABCD is a cyclic quadrilateral in which âˆ BCD = 100Â° and âˆ ABD = 50Â° , find âˆ ADB

Sol.

ABCD is a cyclic quadrilateral,

âˆ A + âˆ C = 180Â° [The opposite angles of a cyclic quadrilateral are supplementary]

â‡’ âˆ A + 100Â°= 180Â°

â‡’ âˆ A = 180Â° – 100Â° = 80Â°

Now in Î” ABD, we have

âˆ A + âˆ ABD + âˆ ADB = 180Â°

â‡’ 80Â° + 50Â° + âˆ ADB = 180Â°

â‡’ âˆ ADB = 180Â° – 130Â° = 50Â°

âˆ´ âˆ ADB = 50Â°

Question 11: In the given figure , O is the centre of the circle and âˆ DAB = 150Â° find value of x and y

Sol.

O is the center of the circle and âˆ BOD = 150Â°

âˆ´ Reflex Â Â âˆ BOD = (360Â° – âˆ BOD)

= (360Â° – 150Â°) = 210Â°

Now, X = Â½(reflex âˆ BOD)

= Â½ Ã— 210Â° = 105Â°

X = 105Â°

Again, X + Y = 180Â°

â‡’ 105Â° + Y = 180Â°

â‡’ Â Â Â Â Â Â Â Â Â Â Y = 180Â° â€“ 105Â° = 75Â°

âˆ´ Â Â Â Â Â Â Â Â Â Â Â Y = 75Â°

Question 12: In the given figure , O is the centre of the circle and âˆ DAB = 50Â° . Find value of x and y.

Sol.

O is the center of the circle and âˆ DAB= 50Â°

OA = OB [Radii]

â‡’ âˆ OBA = âˆ OAB = 50Â°

In Î” OAB, we have

âˆ OAB + âˆ OBA + âˆ AOB =180Â°

â‡’ Â Â 50Â° + 50Â° + âˆ AOB =180Â°

â‡’ Â Â âˆ AOB =180Â° – 100Â° = 80Â°

Since, AOD is a straight line,

X = 180Â° – âˆ AOB

= 180Â° – 80Â° = 100Â°

âˆ´ X = 100Â°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

âˆ DAB + âˆ BCD = 180Â°

âˆ BCD = 180Â° – 50Â° [âˆµ âˆ DAB = 50Â°, given]

â‡’ Y = 130Â°

Thus, X = 100Â° and Y = 130Â°.

Question 13: In the given figure sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If âˆ CBF=130Â° Â and âˆ CDE= X^{0} , find the value of x

Sol.

ABCD is a cyclic quadrilateral

We know that in a cyclic quadrilateral exterior angle = interior angle.

âˆ´ âˆ CBF = âˆ CDA = (180Â° – X)

â‡’ 130Â° = 180Â° – X

â‡’ X = 180Â° – 130Â° = 50Â°

X = 50Â°