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In the adjoining figure, ABCD is a square and PAB is an equilateral triangle. Find
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In the adjoining figure, ABCD is a square and PAB is an equilateral triangle. Find : (i) `angleAPD` (ii) `anglePDC` (iii) `angleDPC` (iv) Prove that `DP = CP`
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jee mains 2020, jee main january, class 12, jee main maths, jee 2020, cbse board, jee main question paper, up board question paper, up boards 2020, up board exam,up board sample paper, cbse class 12, cbse class 10, nageen prakashan, nootan publication
jee mains 2020, jee main january, class 12, jee main maths, jee 2020, cbse board, jee main question paper, up board question paper, up boards 2020, up board exam,up board sample paper, cbse class 12, cbse class 10, nageen prakashan, nootan publication
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