Question

B
14. In the given figure, ABCD is a square and APAB is an equilateral
triangle.
(i) Prove that AAPD = ABPC.
(ii) Show that angleDPC = 15°.
Hint. (i) anglePAD = 90° + 60° = 150° and ZPBC = 90° + 60° = 150°.
(ii) anglePAD = 150° and AP = AD = angleAPD = angleADP = 15°.]
C​

Answer 1

Answer:

Angle PD &HD is

PAD=90

ZPBC=130

Step-by-step explanation:

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