In Fig. 10.24, AB = AC and ACD = 105°, find BAC.

Given,

AB = AC


ACD = 105o


Since, BCD = 180o (Straight angle)


BCA + ACD = 180o


BCA + 105o = 1800


BCA = 75o (i)


Now,


is an isosceles triangle


ABC = ACB (Angle opposite to equal sides)


From (i), we have


ACB = 75o


ABC = ACB = 75o


Sum of interior angle of triangle = 180o


A +B +C =180°


A = 180° - 75° -75°


=30°


Therefore, BAC = 30o


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