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Question
In the given figure, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.
Solution
We can see that the ΔABC is an isosceles triangle with Side AB = Side AC.
⇒ ∠ACB = ∠ABC
As ∠ACB = 65°
hence ∠ABC = 65°
Sum of all the angles of a triangle is 180°
∠ACB + ∠CAB + ∠ABC = 180°
65°+ 65° + ∠CAB = 180°
∠CAB = 180° − 130°
∠CAB = 50°
As BD is parallel to CA
Therefore, ∠CAB = ∠DBA since they are alternate angles.
∠CAB = ∠DBA = 50°
We see that ΔADB is an isosceles triangle with Side AD = Side AB.
⇒ ∠ADB = ∠DBA = 50°
Sum of all the angles of a triangle is 180°
∠ADB + ∠DAB + ∠DBA = 180°
50° + ∠DAB + 50° = 180°
∠DAB = 180° − 100° = 80°
∠DAB = 80°
The angle DAC is the sum of angle DAB and CAB.
∠DAC = ∠CAB + ∠DAB
∠DAC = 50°+ 80°
∠DAC = 130°
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