Question

In the given figure, O is the centre of the circle C1 and AB is the diameter of the circleC2. Quadrilateral PQRS is inscribed in the circle with centre O. Find QRS.
a)45
b)60
c)90
d)135​

Answer 1

Answer:

135

Step-by-step explanation:

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Answer 2

Answer:

R.E.F image

As angle subtended by the diameter on

any point on the circle is 90

∘

,∠AOB=90

(AB diameter in C

2

circle)⇒∠SOQ=90

∘

Also, angle subtended by any arc on the

center is twice to that subtended

on any other point the circle,hence in

circle C

1

,∠SOQ=2×∠QRS (By arc SRQ)

⇒∠QRS=

2

∠SOQ

=

2

90

=45

∘

(15)

By same property

mention above,angles

∠ADC and ∠ABC are equal

(by arc AC)

Now, ∠ADE−∠ABC=15

∘

(∠ADC+∠CDE)−∠ABC=15

∘

⇒(∠ABC+∠CDE)−∠ABC=15

∘

⇒∠CED=15

∘

And angles subtened by are CE,∠CDE=∠CAE

⇒∠CAE=15

∘