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Question
The given figure shows a square ABCD and an equilateral triangle ABP. 
Calculate: (i) ∠AOB
(ii) ∠BPC
(iii) ∠PCD
(iv) Reflex ∠APC
Solution
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°
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