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SHUBHAM SHEKHAR

· started a discussion

· 1 Months ago

आसान सवाल (Question: 111066.0 )

इतना आसान सवाल भी कहीं पूछा जाता है भाई ?

Question:
In the figure, AB = BC = CD = DE = EF = FG = GA, then \(\angle \)DAE is approximately :


Options:
A) 15°
B) 20°
C) 30°
D) 25.7°
Solution:
Ans: (d)

Let \(\angle \)EAD = \(\alpha\) then = \(\alpha\) and also \(\angle \)ACB = \(\alpha\)

Hence \(\angle \)CBD = 2\(\alpha\) (exterior angle to \(\triangle\)ABC),

Since CB = CD, hence \(\angle \)CDB = 2\(\alpha\)


\(\angle \)FGC = 2\(\alpha\) (exterior angle to \(\triangle\)AFG). since GF = EF

\(\angle \)FEG = 2\(\alpha\)

Now \(\angle \)DCE = \(\angle \)DEC = \(\beta\) (say)

Then, \(\angle \)DEF = \(\beta\) - 2\(\alpha\)

Since, \(\angle \)DCB = 180° - (\(\alpha\)+\(\beta\)).

Therefore, in \(\angle \)DCB,

180° - (\(\alpha\)+\(\beta\)) +2\(\alpha\) + 2\(\alpha\) = 180°

or \(\beta\)= 3\(\alpha\)

further \(\angle \)EFD = \(\angle \)EDF = \(\gamma\) (say)

Then, \(\angle \)EDC = \(\gamma\) - 2\(\alpha\)

If CD and EF and in P, then

\(\angle \)FPD = 180° - 5\(\alpha\) (b= 3a)

Now in \(\Delta\)PED, 180° - 5\(\alpha\)+ \(\gamma\) +2\(\alpha\) = 180° or \(\beta\) = 3\(\alpha\)

Therefore in \(\Delta\)EFD   

\(\alpha\) +2\(\gamma\)= 180° or \(\alpha\) + 6\(\alpha\) = 180°

or \(\alpha\) =25.7° 

Knowledge Expert

· commented

· 1 Months ago

Dear Student,
The question is neither tuff nor easy.
It is tricky one.
You can solve it using formula also. 180 / n , where n is no of equal given sides.

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