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Question
In Figure AB = AC and ∠ACD =105°, find ∠BAC.
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Solution
Consider the given figure
We have,
AB = AC and ∠ACD =105^@
Since,
∠BCD = 180° = Straight angle
∠BCA + ∠ACD = 180°
∠BCA +105° = 180°
∠BCA = 180° -105° ⇒ ∠BCA = 75° .......1
And also,
ΔABC is an isosceles triangle [∵AB = AC]
⇒ ∠ABC = ∠ACB
From (1), we have [Angles opposite to equal sides are equal]
∠ACB = 75° ⇒∠ABC = ∠ACB = 75°
And also,
Sum of interior angles of a triangle = 180°
⇒∠ABC = ∠BCA + ∠CAB = 180°
⇒ 75° + 75° + ÐCAB = 180°
⇒ 150° +∠BAC = 180°⇒ ∠BAC = 180° -150° = 30°
∠BAC=30°
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