Question

In the figure given in the attachment, PA and PB are tangents at the points A and B respectively of a circle with centre O. Q and R are points on the circle. If ZAPB = 70°, find :
(i) ∠AOB (ii) ∠AQB (iii) ∠ARB.
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Answer 1

Step-by-step explanation:

Given PA & PB are tangent to the circle with center O. PA = PB [length of tangent from external point to circle are equal] .(1) In ΔΡ ΑΒ PA PB ZPBA = ZP AB [isosceles triangle] now ZP AB+ZPBA+ZAPB = 180° [Angle sum prop] 2/P AB 180 - 50 = 130 ZPBA ZP AB = 65° Now PA is tangent & OA is radius at point A. ZOAP 90° [tangent at any point is to = Σ radius] ZOAB = ZOAP - ZP AB= 90- 65 = 25° Hence angle OAB is 25º.

Sorry little mistake I have done in this step by step explaining instead of 70 I have wrote 50 so pls change it and you will get the correct answer bro

Just change 50 into 70 and then minus answer you will get the result

Answer 2

Hint: Recall the properties of Tangent Lines and the Angle at the Center Theorem of Circles.

Solution:-

Tangent lines are always perpendicular to the radius of the circle at the point of tangency i.e. the angle made by tangent with radius is always 90° at the point where tangent touches the circle.

By using this property of tangent lines, we get:

• ∠PAO = 90°
• ∠PBO = 90°

... since PA and PB are tangents and OA and OB are the radius of circle respectively.

Apply angle sum property of quadrilateral in PAOB.

⇒ ∠PAO + ∠PBO + ∠AOB + ∠APB = 360°

⇒ 90° + 90° + ∠AOB + 70° = 360°

⇒ ∠AOB = 360° - 70°- 90° - 90°

⇒ ∠AOB = 110°

Now, we will use a theorem of circle according to which, the angle at center,. created by an arc is 2 times the angle at circumference.

By using this property, we get:

• ∠AOB = 2 ∠AQB

Now substitute the value of know angle.

⇒ ∠AOB = 2 ∠AQB

⇒ 110° = 2 ∠AQB

⇒ 110°/2 = ∠AQB

⇒ 55° = ∠AQB

To find the value of ∠ARB, we will use the same theorem, but for the implementation, we have to find the value of obtuse ∠AOB.

We know that a complete revolution is of 360°. By using this concept, we get the following results:

• Acute ∠AOB + Obtuse ∠AOB = 360°

Now, substitute the value of acute ∠AOB which we have already obtained in (i) question.

⇒ Acute ∠AOB + Obtuse ∠AOB = 360°

⇒ 110° + obtuse ∠AOB = 360°

⇒ obtuse ∠AOB = 360° - 110°

⇒ obtuse ∠AOB = 250°

Now, according to the theorem following result holds true:

• obtuse ∠AOB = 2 ∠ARB

Substitute the value of obtuse ∠AOB.

⇒ obtuse ∠AOB = 2 ∠ARB

⇒ 250° = 2 ∠ARB

⇒ 250°/2 = ∠ARB

⇒ 125° = ∠ARB

Final Answers:-

(i) ∠AOB = 110°

(ii) ∠AQB = 55°

(iii) ∠ARB = 125°