How do I find x?

All triangles add up to 180 degrees so in triangle ACD you add 42 to 90 and you get 132 then you subtract 180 to 132 and you'll get 48. So now you need to find z and y so at angle c it's a right angle so now you need to subtract 90 from 48 and you'll get 42. Now that you found z all you have is y to find y you need to add 90 to 42 and again you get 132. Then subtract to 180 and you get 48.

x = 90-42 = 48 deg
z = 42 deg
y = 48 deg

x+42 deg = 90 deg (exterior angle of triangle)
x = 48 deg

z = 90 deg - x

y + z = 90 deg (exterior angle of triangle)
y + (90 deg - x) = 90 deg
y + 90 deg - x = 90 deg
y - x = 0
y = x
y = 48 deg

no

It's very easy to find x
∵ ∠CDA=90°
∠ACB=90°
∠CAB=42°
∴X=∠CDA-∠CAB=90°-42°=48°
∴Z=48°
y=42°

Oh crap...I'm taking this in a few semesters -_-

Good luck!

Look at similar triangles ADC and CDB. y is not 42 degrees.

I ll help you. So, the angle opposite the square which is 90•, so that will be 90• as well. So 90+42=132 180-132=148 so x = 148• now try and work out y

sum of angles in triangle = 180

Because all angles in a triangle must add up to 180, 42 + 90 + x = 180. (I'm using triangle ACD.) You do the math.

Also, yes, y should also be 42.