If tangents AP and PB from a point to a circle with centre O are inclined to each other at 80 degrees then angle POA is equal to?

Question

Answer

Answer

Angle POA = 50 degree.

Answer

Please someone answer....

Answer

Given
PA=PB
therefore
angle P=80'
Angle A+ Angle P+ Angle O= 180' ( angle sum property)
90'+40'+ angle is equal to 180
180 - 130 is equal to 50
therefore angle POA =50

Answer

Given information:
- Tangents AP and PB are drawn from a point P to a circle with center O.
- The angle between tangents AP and PB is 80 degrees.

To find:
The measure of angle POA.

Explanation:
Step 1: Understanding the problem
We have a circle with center O and a point P outside the circle from which two tangents AP and PB are drawn. We need to find the measure of angle POA.

Step 2: Analyzing the given information
From the given information, we know that tangents AP and PB are inclined to each other at 80 degrees. Let's denote the intersection point of tangents AP and PB as point X.

Step 3: Identifying key points and angles
- Point O: Center of the circle
- Point P: External point from which tangents are drawn
- Point A: Point of tangency between tangent AP and the circle
- Point B: Point of tangency between tangent PB and the circle
- Point X: Intersection of tangents AP and PB
- Angle APB: Angle between tangents AP and PB (given as 80 degrees)
- Angle POA: Angle we need to find

Step 4: Drawing a diagram
Let's draw a diagram to visualize the given information and the points mentioned above.

Step 5: Applying tangent properties

The tangent lines drawn from an external point to a circle are equal in length. Therefore, we can conclude that:
AP = BP

Step 6: Applying the property of cyclic quadrilaterals

Since the points A, B, P, and X lie on the circle, we have a cyclic quadrilateral. The opposite angles of a cyclic quadrilateral are supplementary. Therefore, we can conclude that:
Angle APB + Angle AXB = 180 degrees

Since we know that Angle APB is 80 degrees, we can substitute it into the equation above to find:
80 + Angle AXB = 180 degrees

Solving for Angle AXB, we get:
Angle AXB = 100 degrees

Step 7: Finding angle AOX

Since Angle AXB is an external angle of triangle AOX, it is equal to the sum of the two opposite interior angles of the triangle. Therefore, we have:
Angle AOX + Angle AXO = Angle AXB

Since Angle AXB is 100 degrees, and we know that Angle AXO is a right angle (90 degrees), we can substitute these values into the equation above to find:
Angle AOX + 90 = 100 degrees

Solving for Angle AOX, we get:
Angle AOX = 10 degrees

Step 8: Finding angle POA

Angle POA is the sum of angles AOX and AOP. Since we know that Angle AOX is 10 degrees, we need to find Angle AOP.

Since AO is the radius of the circle and AP is a tangent, Angle OAP is a right angle (90 degrees). Therefore, Angle AOP is equal to the complement of Angle OAP. We can conclude that:
Angle AOP = 90 - Angle OAP

Since Angle OAP and Angle AOX are vertically opposite angles, they are equal

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