Question:
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°
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