+0  
 
0
104
3
avatar

In the figure shown, $ABCD$ is a square and $\triangle CDE$ is equilateral. What is the degree measure of $\angle CBE$?

l

 Jan 31, 2021
 #1
avatar
0

By the Cosine Law, BE  = 1.8794, and

cos angle EBC = (BE^2 + BC^2 - CE^2)/(2*BE*BC) = cos 20, so angle EBC = 20 degrees.

 Jan 31, 2021
 #2
avatar+536 
0

Triangle CBE is isosceles because the square side length is equal to the triangle side length.

TO find the remaining angles, we find the middle angle, and the other two will be half because it is isosceles.

Angle ACB is equal to the squares interior angle plus the interior of the triangle.

90+60=150.

The remaining 2 angles are equal to each other.

So 180-150=30.

30/2=15.

 Jan 31, 2021
 #3
avatar
0

Thank you!

 Jan 31, 2021

27 Online Users

avatar
avatar
avatar
avatar