In a kite ABCD, AB=CB , AD = CD, <B= 115 degree, <D=30 degree. Find <A = ?

The measure of ∠A = 107.5°

Step-by-step explanation:

Given that

ABCD is a kite such that

AB = BC and AD = DC

Join A and C

Since AB = BC

→ ∠BAC = ∠BCA

Since , the angles opposite to equal sides are equal.

Let ∠BAC = ∠BCA = x°

and

Since AD = DC

→ ∠CAD = ∠ACD

Since , the angles opposite to equal sides are equal.

Let ∠CAD = ∠ACD = y°

Now

In ∆ ADC ,

∠ADC + ∠CAD + ∠ACD = 180°

Since , the sum of the all angles in a triangle is 180°

⇛ 30° + y°+ y° = 180°

⇛ 30°+2y° = 180°

⇛ 2y° = 180°-30°

⇛ 2y° = 150°

⇛ y° = 150°/2

⇛ y° = 75°

Therefore, ∠CAD = ∠ACD = 75°

and

In ∆ ABC,

∠ ABC + ∠BAC + ∠BCA = 180°

Since , the sum of the all angles in a triangle is 180°

⇛ 115° + x° + x° = 180°

⇛ 115°+2x° = 180°

⇛ 2x° = 180°-115°

⇛ 2x° = 65°

⇛ x° = 65°/2

⇛ x° = 32.5°

Therefore, ∠BAC = ∠BCA = 32.5°

Now, ∠ A = ∠BAC + ∠CAD

⇛ ∠A = 32.5° + 75°

⇛ ∠A = 107.5°

Therefore, ∠A = 107.5°

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