Mathematics‘O’ is the center of the circle, AC is a tangent to a circle at point A. If ΔOAC  is isosceles triangle then find the measure of ∠OCA

'O' is the center of the circle, AC is a tangent to a circle at point A. If ΔOAC  is isosceles triangle then find the measure of OCA


  1. A
    75 °  
  2. B
    50 °  
  3. C
    45 °  
  4. D
    40 °   

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    Solution:

    Let us first draw diagram for better understanding:
    IMG_256
    Let O be the center of the circle and let AC be the tangent to the circle. Join OA which is the radius of the circle. Also the joint OC which forms the triangle OAC.
    As we know, the radius of a circle is perpendicular to the tangent of the circle. Therefore
    OAC= 90 °  .
    Now, we have to find  OCA   let us suppose that  OCA=x  
    As we are given that  ΔOCA   is an isosceles triangle. Therefore, OA = AC
    As we know, opposite angles of equal sides are also equal by isosceles triangle property. Therefore, AOC=OCA  .
    As we have supposed OCA=x   and AOC=x  
    Hence, in ΔOAC,OAC= 90 ,OCA=x and AOC=x   OAC+OCA+AOC= 180 °    (sum of angles of triangle property)
    Putting values of angles found earlier, we get :
    90 ° +x+x= 180 ° 90 ° +2x= 180 ° 2x= 90 ° x= 45 °  
    But  OCA=x  
    Hence,  OCA= 45 °  which is our required angle.

    So, the correct Ans is Option 3.
     
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