Question

In the following figure, ∠ PQR = 100°, P, Q and R are points on a circle with centre O. Find ∠OPR.​

Answer 1

Answer:

Here, PR is chord

We mark s on major arc of the circle.

∴ PQRS is a cyclic quadrilateral.

So, ∠PQR+∠PSR=180o

[Sum of opposite angles of a cyclic quadrilateral is 180o]

100+∠PSR=180o

∠PSR=180o−100o

∠PSR=80o

Arc PQR subtends ∠PQR at centre of a circle.

And ∠PSR on point s.

So, ∠POR=2∠PSR

[Angle subtended by arc at the centre is double the angle subtended by it any other point]

∠POR=2×80o=160o

Now,

In ΔOPR,

OP=OR[Radii of same circle are equal]

∴∠OPR=∠ORP [opp. angles to equal sides are equal] ………………..(1)

Also in ΔOPR,

∠OPR+∠ORP+∠POR=180o (Angle sum property of triangle)

∠OPR+∠OPR+∠POR=180o from (1)

2∠OPR+160=180o

2∠OPR=180o−160o

2∠OPR=20

∠OPR=20/2

∴∠OPR=10o.

Step-by-step explanation:

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Answer 2

Answer:

Solution:

Since angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.

So, the reflex ∠POR = 2 × ∠PQR

We know the values of angle PQR as 100°

So, ∠POR = 2 × 100° = 200°

∴ ∠POR = 360° – 200° = 160°

Now, in ΔOPR,

OP and OR are the radii of the circle

So, OP = OR

Also, ∠OPR = ∠ORP

Now, we know sum of the angles in a triangle is equal to 180 degrees

So,

∠POR + ∠OPR + ∠ORP = 180°

⇒ ∠OPR + ∠OPR = 180° – 160°

As ∠OPR = ∠ORP

⇒ 2∠OPR = 20°

Thus, ∠OPR = 10°

I HOPE IT CAN HELP YOU FRIEND(✿◡‿◡)

Step-by-step explanation: