Question

In the given figure, AB and XY are diameters of a circle
with centre O. If <APX = 30°, find :
(i) <AOX
(ii) <APY
(iii) <BPY
(iv) <OAX.​

Answer 1

Given:

AB and XY are diameters of a circle  with centre O

∠APX = 30°

To find:

(i) ∠AOX

(ii) ∠APY

(iii) ∠BPY

(iv) ∠OAX

Solution:

(i). Finding the measure of ∠AOX:

We know that,

The angle subtended by an arc of a circle at its centre is double the angle it subtends anywhere on the circle's circumference .

So, we get

∠AOX = 2 ∠APX

substituting ∠APX = 30°, we get

⇒ ∠AOX = 2 × 30°

⇒ ∠AOX = 60°

Thus, $\boxed{\bold{\angle AOX = \underline{60\°}}}$

(ii). Finding the measure of ∠APY:

XY is given as a diameter

We know that → The angle subtended on a semi-circle is a right angle.

∴ ∠XPY = 90°

From the attached figure, we get

∠APY = ∠XPY - ∠APX = 90° - 30° = 60°

Thus, $\boxed{\bold{\angle APY= \underline{60\°}}}$.

(iii). Finding the measure of ∠BPY:

AB is given as a diameter

We know that → The angle subtended on a semi-circle is a right angle.

∴ ∠APB = 90°

From the attached figure, we get

∠BPY = ∠APB - ∠APY = 90° - 60° = 30°

Thus, $\boxed{\bold{\angle BPY= \underline{30\°}}}$.

(iv). Finding the measure of ∠OAX:

In Δ AOX,

OA = OX ...... [radius of the circle]

∴ ∠OXA = ∠OAX ..... (1) .... [angles opposite to equal sides are also equal]

Now,

∠OXA + ∠OAX + ∠AOX = 180° ...... [Angle sum property of a triangle]

substituting from (1) and the value of ∠AOX = 60°, we get

⇒ 2∠OAX + 60° = 180°

⇒ 2∠OAX  = 180° - 60°

⇒ 2∠OAX  = 120°

⇒ ∠OAX  = $\frac{120\°}{2}$

⇒ ∠OAX = 60°

Thus, $\boxed{\bold{\angle OAX = \underline{60\°}}}$.

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Answer 2

Step-by-step explanation:

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