The given figures shows a kite-shaped figure whose diagonals intersect each other at point O. If ∠ABO = 25° and ∠OCD = 40°; find

(i) ∠ABC

(ii) ∠ADC

(iii) ∠BAD

Given: ABCD is a kite, so, AD = DC and AB = BC. The diagonals of the kite intersect at point O.

∠ABO = 25° and ∠OCD = 40°

(i) In triangle AOB, we know that the diagonals of the kite are perpendicular to each other at point O.

⇒ ∠AOB = 90°

Sum of all angles in triangle is 180°.

⇒ ∠AOB + ∠ABO + ∠BAO = 180°

⇒ 90° + 25° + ∠BAO = 180°

⇒ 115° + ∠BAO = 180°

⇒ ∠BAO = 180° - 115°

⇒ ∠BAO = 65°

Since AB = AC (because ABCD is a kite), we have:

⇒ ∠BCO = ∠BAO = 65°

In triangle ABC, the sum of the angles is 180°.

⇒ ∠ABC + ∠BCA + ∠BAC = 180°

⇒ ∠ABC + 65° + 65° = 180°

⇒ ∠ABC + 130° = 180°

⇒ ∠ABC = 180° - 130°

⇒ ∠ABC = 50°

Hence, ∠ABC = 50°.

(ii) In triangle ADC, we know that AD = DC (because ABCD is a kite), so the angles opposite these equal sides are equal.

⇒ ∠DAC = ∠ACD

We are given that ∠OCD = 40° and since the diagonals bisect the angles at C, we have:

∠ACD = ∠OCD = 40°

Now, using the sum of angles in triangle ADC:

⇒ ∠DAC + ∠ACD + ∠ADC = 180°

⇒ 40° + 40° + ∠ADC = 180°

⇒ 80° + ∠ADC = 180°

⇒ ∠ADC = 180° - 80°

⇒ ∠ADC = 100°

Hence, ∠ADC = 100°.

(iii) In triangle BAD,

∠BAD = ∠BAO + ∠OAD

From earlier, we found: ∠BAO = 65°

Since AD = DC (because ABCD is a kite), the angle bisector of ∠ADC divides it into two equal parts:

∠OAD = ∠ACD = 40°

∠BAD = 65° + 40° = 105°

Hence, ∠BAD = 105°.