Refer to the notations above:
<BCD+40°=180° (angles on a strainght line)
<BCD=180°-40°
<BCD=140°
As BC=CD(given),△BCD is an isosceles triangle,thus <CBD=<CDB=x(properties of isosceles triangle).
We can then proceed to find x(<CDB and <CBD):
140°+ x + x=180°(angles sum of triangles)
2x=180°-140°
x=40°/2
x=20°
Then, we can again use the angles sum of triangle to solve y:
y+90°+20°=180°
y=180°-90°-20°
y=70°
Therefore, x=20° and y=70°.
Hope it helps!
