A) \[60{}^\circ ,\text{ }120{}^\circ \]
B) \[120{}^\circ ,\text{ }60{}^\circ \]
C) \[\text{80 }\!\!{}^\circ\!\!\text{ , 100 }\!\!{}^\circ\!\!\text{ }\]
D) \[100{}^\circ ,\text{ }80{}^\circ \]
Correct Answer: A
Solution :
In the given figure, ABCD is a rhombus and \[DP\_AB\] such that \[AP=PB.\] Join BD. Let AB = BC = CD = AD = 2a So, AP = a = PB In \[\Delta APD\], \[D{{P}^{2}}=A{{D}^{2}}-A{{P}^{2}}\] (Pythagoras theorem) =\[{{(2a)}^{2}}-{{a}^{2}}=4{{a}^{2}}-{{a}^{2}}=3{{a}^{2}}\] In \[\Delta DPB\]we have \[B{{D}^{2}}=D{{P}^{2}}+P{{B}^{2}}=3{{a}^{2}}+{{a}^{2}}=4{{a}^{2}}\] \[\therefore BD=2a\] So, \[AB=BO=AD=2a\] \[\therefore \Delta ABD\]is an equilateral triangle. \[\therefore \angle A=60{}^\circ \] and \[\angle ABD=60{}^\circ \] Similarly, \[\Delta \text{BDC}\]is an equilateral triangle and \[\angle \text{DBC=60 }\!\!{}^\circ\!\!\text{ }\] Now, \[\angle ABC=\angle ABD+\angle DBC=60{}^\circ +60{}^\circ =120{}^\circ \] So, \[\angle A=60{}^\circ \text{ and }\angle \text{B=120}{}^\circ \].
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