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Question

In the given figure, O is the centre of the circumcircle. Tangents at A and C intersect at P. Given AOB=140 and APC = 80; find the BAC.

  1. 90
  2. 20
  3. 60
  4. 50


Solution

The correct option is C 60
Join OC 
Given: AOB = 140
In OAB,
OA = OB     [ radius of circle]
By Theorem- Angles opposite to equal sides of a triangle are equal.
OAB =  OBA     ---- (i)

By Theorem- Sum of angles of a triangle = 180
  OAB + OBA +  AOB = 180
2 OAB + 140180     [from (i)]

OAB  = (180140)2
OAB = 20

By Theorem- If two tangents PA and PC are drawn to a circle with centre O from an external point P, then APC = 2 OAC = 2 OCA

OAC = 12  APC
Given:  APC = 80
  OAC =  802
  OAC =  40

Now,
The required BAC = OAB + OAC
 BAC = 2040
 BAC = 60

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