question_answer

In an equilateral triangle \[{{45}^{o}}\] then:        

A)  \[{{90}^{o}}\]        

B)  \[\angle A=\angle B=\angle C=\angle D={{90}^{o}}\]       

C)  \[x<y\]     

D)  \[x>y\]       

Correct Answer: C

Solution :

Let ABC be an equilateral \[{{105}^{o}}\] and \[{{85}^{o}}\]. Since, A ABD is a right angled \[{{60}^{o}}\]\[\angle BAC={{45}^{o}}\], right angle at D. \[\angle BED={{120}^{o}}\] \[\angle ABD\] (By Pythagorus theorem.) \[{{15}^{o}}\]  \[{{30}^{o}}\]      (\[\because BD=BD\]) \[\angle CAD={{40}^{o}}\]   \[\angle BDC={{25}^{o}}\] \[\angle BDC\]       (\[{{85}^{o}}\]\[{{120}^{o}}\]) \[{{115}^{o}}\] \[{{95}^{o}}\] \[\angle PAD={{30}^{o}}\] \[\angle CPA={{45}^{o}}\]