Question

in the given fig pt o is the centre of circle find the value of angle ACP if angle POB is 90 °​

Answer 1

Answer:

Step-by-step explanation:

We have angle APB = 75°  

Now , PA and PB are the tangents to the circle from P at A and B respectively and OA and OB are the radii of the circle.  

We know that tangent to a circle is always _I_ to the radius at the point of contact.

 

Hence, OA _|_ PA  and OB_|_ PB  

Now, angle OAP = angle OBP = 90°  

In quadrilateral AOBP  

Angle ABP + Angle OAP + angle OBP + angle AOB = 360°  

= 75° + 90° + 90° + Angle AOB = 360°  

= 255° + angle AOB = 360°  

Angle AOB = 360° - 255°  

Angle AOB = 105°  

We know that angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle.  

Consider the minor arc AB, then  

Angle AOB = 2 angle AQB  

= angle AQB= ½ angle AOB = ½ x 105°  

= angle AQB = 52.5°  

Since AQBM is a cyclic quadrilateral, then  

Angle AMB + angle AQB = 180° [opposite angles of cyclic quad are supplementary]  

Angle AMB + 52.5° = 180°  

Angle AMB = 180° - 52.5°  

Angle AMB = 127.5°