Question

in the given alongside figure AD=AB=AC,BD parallel to CAand angle ACB=65 degree find angle DAC

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Answer 1

where's the figure????????

Answer 2

∠DAC = 130°

Step-by-step explanation:

From the figure in attachment,

Let AC = AB = AD = a

Given ∠ACB = 65°

ΔABC is an isosceles triangle with sides AB = AC.

∴ ∠ABC = ∠ACB = 65° (Given)

∵ Sum of all angles in a triangle is 180°,

∠BAC + ∠ABC + ∠ACB = 180°

∴ ∠BAC = 50°.

Now, since AC is parallel to BD

∠DBA = ∠BAC

∴ ∠DBA = 50°

Since, ΔDAB is also an isosceles triangle,

∠ADB = ∠ABD = 50°

Since the sum of angles in a triangle is 180°

∠DAB = 80°

∠DAC = ∠DAB + ∠BAC (from figure)

∠DAC = 80° + 50° = 130°