Step: 1

Given, ∠ OQR = 22^{o}.

Step: 2

Step: 3

The angle made by the chord QR at the center of the circle is twice the angle made by it at any point on the circle.

Step: 4

So, m ∠ QPR = 136/2 = 68°.

Correct Answer is : 68

Step: 1

Quadrilateral PQRS is a cyclic quadrilateral.

Step: 2

Therefore,∠ ASR = ∠ PQR

Step: 3

From the given choices, ∠ Q=80°

Correct Answer is : ∠Q = 80°

Step: 1

[Sum of the angles in a triangle is 180°.]

Step: 2

49° + 70° + m ∠BAC = 180°

Step: 3

Step: 4

Join BO and CO.

Step: 5

[Inscribed angle theorem.]

Step: 6

Therefore, m ∠ BOC = 122° .

Correct Answer is : 122°

Step: 1

[Given.]

Step: 2

Join A C and BO .

Step: 3

[Inscribed angle theorem.]

Step: 4

[Sum of the angles in a triangle is 180°.]

Step: 5

90° + m ∠ACB + 37° = 180°

Step: 6

Step: 7

[Inscribed angle theorem.]

Step: 8

[Substitute.]

Step: 9

Therefore, x = 106°.

Correct Answer is : 106°

Step: 1

[Given.]

Step: 2

Join B D .

Step: 3

[Inscribed angle theorem.]

Step: 4

m∠ ABC = m∠ ABD + m∠ CBD

[Angle Addition Postulate.]

Step: 5

114° = 84° + m∠ CBD

[Substitute.]

Step: 6

m∠ CBD = 114° - 84° = 30°

Step: 7

Also, m ∠ABD + m ∠ADB + m ∠BAD = 180°

[Sum of the angles in a triangle is 180°.]

Step: 8

84° + m ∠ADB + 65° = 180°

Step: 9

Step: 10

Join A O and B O .

Step: 11

[Inscribed angle theorem.]

Step: 12

Join C O and D O .

Step: 13

[Inscribed angle theorem.]

Step: 14

Now, m ∠AOD + m ∠AOB + m ∠BOC + m ∠COD = 360°

[The sum of angles around a point will always be 360°.]

Step: 15

168° + 62° + x + 60° = 360°

[Substitute.]

Step: 16

So, x = 360° - 290° = 70°

Correct Answer is : 70°

Step: 1

[Inscribed angle theorem.]

Step: 2

(3x + 2)° = 1 2 × 196° = 98°

[Substitute.]

Step: 3

3x = 98 - 2 = 96

Step: 4

Therefore, x = 96 3 = 32.

Correct Answer is : 32

Step: 1

Join B O .

Step: 2

[Given]

Step: 3

[Linear pair.]

Step: 4

(7x - 10)° + (3x - 10)° = 180°

[Substitute.]

Step: 5

10x - 20 = 180

Step: 6

10x = 200 ⇒ x = 200 10 = 20

Step: 7

m∠ BOC = (3x - 10)° = (3(20) - 10)° = 50°

Step: 8

[Inscribed angle theorem.]

Correct Answer is : 25°

Step: 1

[Given]

Step: 2

The opposite angles of a quadrilateral inscribed in a circle are supplementary.

Step: 3

[ABCD is a cyclic quadrilateral.]

Step: 4

[Substitute and simplify.]

Step: 5

[Angle in a semi circle.]

Step: 6

In ∠ADB, m ∠ DAB + m ∠ ABD + m ∠ADB = 180°

[Triangle Angle- Sum Theorem.]

Step: 7

Step: 8

x° = 180° - (75° + 90°) = 180° - 165° = 15°

[Substitute and simplify.]

Step: 9

Therefore, x = 15.

Correct Answer is : 15

- Inscribed Angle