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