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Q.

'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


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a

75 °  

b

50 °  

c

45 °  

d

40 °   

answer is C.

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Detailed 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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