Question

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

Calculate: (i) ∠AOB
                (ii) ∠BPC
                (iii) ∠PCD
                (iv) Reflex ∠APC

Answer 1

In the given figure ΔAPB is an equilateral triangle.
Therefore all its angles are 60°
Again in the
ΔADB,
∠ABD = 45°
∠AOB = 180° - 60° - 45° = 75°

Again
ΔBPC
⇒ ∠BPC = 75°              ....[ Since BP = CB ]
Now,
∠C = ∠BCP + ∠PCD 
⇒ ∠PCD = 90° - 75°
⇒ ∠PCD = 15°
Therefore,
∠APC = 60° + 75°
⇒ ∠APC = 135°
⇒ Reflex ∠APC = 360° - 135° = 225°

(i) ∠AOB = 75°
(ii) ∠BPC = 75°
(iii) ∠PCD = 15°
(iv) Reflex ∠APC = 225°