shridhar786 wrote:
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In the diagram below an equilateral triangle, CDE has been inscribed inside a square ABCD. What is the value of the angle marked x°?
(A) 15
(B) 30
(C) 45
(D) 75
(E) it can't be determined from the facts given
I have a query.
Refer the figure below.
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File comment: Equilateral Triangle DEC
Equilateral Triangle DEC.png [ 18.12 KiB | Viewed 392 times ]
In the figure DE = EC = CD = a. EF is perpendicular to AD and EG is perpendicular to DC.
Angle DEF = 60°. EG = DF = \(\frac{\sqrt{3}}{2}*a\). AF = \(a - \frac{\sqrt{3}}{2}*a\).
Since EG is perpendicular to DC, G bisects DC.
Hence DG = FE = \(\frac{a}{2}\)
Now, in triangle AEF, angle AEF = y
Thus, Tan y = AF/EF = \((a - \frac{\sqrt{3}}{2}*a)/\frac{a}{2}\)
Tan y = 1 - \(\frac{\sqrt{3}}{2}\)
Till this point it was clear that I need to know trigonometry to find 'y' but it was difficult to find exact value.
Since n know the angle an be found using trigonometry and it is greater than 60°, chose D(E is not possible).
Is it right to do this way.?
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