Question

Page : 7.
Answer Any two or all three but 27th is Compulsory.

Thanks :-)

Answer 1

Hey mate !!!

here is your answer !!!

< RQU = <ASR = 65° { angle exterior of cyclic Quadrilateral }




< ASR = 65°


now , <AOR = 2<ASR

=> <AOR = 65° x 2

< AOR = 130

NOW ,

<AOR + <Y = 360 ° ( COMPLETE angle )

130° + <Y = 360°

Hence , y = 230°


hence , Option number C is correct !!!


hope it helps you !!

thanks

Answer 2

Hey friend, Harish here.

Here is your answer:

27) 

 Given that,

1)   ∠RQU = 65°

To find,

∠ y.

Solution:

We know that,

∠RQT + ∠RQU = 180°  ( Straight line)

⇒ ∠RQT + 65° = 180°

⇒ ∠RQT = 180° - 65° = 115°.

Now, We know that, SRQA is an cyclic quadrilateral because it's vertices touches the circle.

Then,

∠S + ∠RQT = 180°  (Sum of opp angles of cyclic quadrilateral is 180°)

⇒ ∠S + 115° = 180° 

⇒ ∠S = 180° - 115° = 65°.

We also know that, 

2 × ∠S = ∠AOR  (Angle made at center is twice the angle made at any other point touching the circle).

∴ ∠AOR = 2 × 65 = 130°.

We know that, 

∠AOR + ∠Y = 360° ( whole circle angle)

⇒ ∠y + 130 = 360°

⇒ ∠y = 360 - 130 = 230°

Therefore angle y is 230°. (OPTION - C)
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Hope my answer is helpful to you.