The given figure shows a square ABCD and an equilateral triangle ABP. Calculate :

(i) ∠AOB

(ii) ∠BPC

(iii) ∠PCD

(iv) reflex ∠APC

(i) Given,

ABP is an equilateral triangle.

∴ ∠PAB = 60°

From figure,

⇒ ∠OAB = ∠PAB = 60°.

We know that,

Each interior angle of a square equals 90° and diagonals bisect interior angles.

∴ ∠DBA = $\dfrac{90°}{2}$ = 45°.

From figure,

⇒ ∠OBA = ∠DBA = 45°.

In △ AOB,

By angle sum property of triangle,

⇒ ∠OBA + ∠OAB + ∠AOB = 180°

⇒ 45° + 60° + ∠AOB = 180°

⇒ ∠AOB + 105° = 180°

⇒ ∠AOB = 180° - 105° = 75°.

Hence, ∠AOB = 75°.

(ii) From figure,

⇒ ∠PBA = 60° [Each angle of an equilateral triangle equals to 60°.]

⇒ ∠CBP = ∠CBA - ∠PBA = 90° - 60° = 30°.

We know that,

⇒ BP = AB (Sides of equilateral triangle) ………(1)

⇒ AB = BC (Sides of square ABCD are equal) ……..(2)

From equation (1) and (2), we get :

⇒ BP = BC.

In △ BPC,

⇒ BP = BC

⇒ ∠BCP = ∠BPC = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠BCP + ∠BPC + ∠CBP = 180°

⇒ x + x + 30° = 180°

⇒ 2x = 180° - 30°

⇒ 2x = 150°

⇒ x = $\dfrac{150°}{2}$ = 75°.

Hence, ∠BPC = 75°.

(iii) As,

⇒ ∠BCP = ∠BPC = 75°

From figure,

⇒ ∠C = ∠BCP + ∠PCD

⇒ 90° = 75° + ∠PCD

⇒ ∠PCD = 90° - 75° = 15°.

Hence, ∠PCD = 15°.

(iv) From figure,

⇒ ∠APC = ∠APB + ∠BPC

⇒ ∠APC = 60° + 75° = 135°

⇒ Reflex ∠APC = 360° - ∠APC = 360° - 135° = 225°.

Hence, reflex ∠APC = 225°.