In a ∆ABC = AC and ∠ACD = (10)5°. Find ∠BAC.

Solution:

We have,

AB = AC and ∠ACD = (10)5°

Since, ∠BCD = 180° = Straight angle

∠BCA + ∠ ACD = 180°

∠BCA + (10)5° = 180°

∠BCA = l80° - (10)5°

∠BCA = 75°

And also,

ΔABC is an isosceles triangle      [AB = AC]

∠ABC = ∠ ACB [Angles opposite to equal sides are equal]

From (i), we have

∠ACB = 75°

∠ABC = ∠ACB = 75°

And also,

Sum of Interior angles of a triangle = 180°

∠ABC = ∠BCA + ∠CAB = 180°

75° + 75° + ∠CAB =180°

150° + ∠BAC = 180°

∠BAC = 180° - 150° = 30°

∠BAC = 30°