In an triangle, AC = BC and #AC^2 = 2AB^2#, then what will be angle of C?

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1 Answer
CW
Dec 25, 2016

#41.41^@#

Explanation:

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Given #AC=BC, => DeltaABC# is isosceles.
Given #AC^2=2AB^2#,
#AC^2=AD^2+DC^2#
#AD=1/2AB, => AD^2=1/4AB^2#
#=> 2AB^2=1/4AB^2+DC^2#
#=> 7/4AB^2=DC^2, => DC=sqrt(7/4)AB=sqrt7/2AB#
#tanx=(AD)/(DC)=(1/2AB)/(sqrt7/2AB)#
#tanx=1/sqrt7#
#=> x=tan^-1(1/sqrt7)=20.705^@#

#=> angleACB=angleC=2x=2*20.705=41.41^@#

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