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In square $ABDE$, $\triangle CDE$ is isosceles with $\angle CED = \angle CDE = 15^\circ$.  Find the measure of $\angle ACB$ in degrees.

 

I think the answer is 60, but how do I prove it?

 

 Jan 24, 2021
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The angles in a triangle is 180 degrees so 180-15x2=150.  The angle of interior and exterior is 360 so 360-150=210.  Angle BDC is 75 degrees  because it is a box so 90 - 15=75.  75 is for both sides so that means 210 - 150= 60 degrees.  That is proof that 60 is the correct answer.

 Jan 24, 2021

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