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Question

In the given fig. ABCD is a square and PAB is an equilateral triangle.
(i) Prove that APDBPC
(ii) Show that DPC=30
1023985_28f473db908a4f4092c97e0ba2ade0cf.png

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Solution

In ABP,
AP=BP [Sides of equilateral ]
& AD=BC [Sides of equilateral ]
and DAP=DABPAB=9060=30
Now in APD and BPC
AP=BP [Sides of equilateral ]
AD=BC [Sides of equilateral ]
DAP=CBP [Both 30]
APDBPC [By SAS]
In APD
AP=AD [ AP=AB ]
DAP=30
APD=180302=75
Similarly, BPC=75
Therefore, DPC=360(75+75+60)=150

968153_1023985_ans_09400972bd8c4a8eb889693f2dcdc2cf.png

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