In ΔABC, m∠ACB = 90°, CD ⊥ AB and m∠ACD = 45°. Find: CD, if BC = 3 in

We can find m∠BCD like follows: m∠BCD=90°-45°=45°
Now, m∠DBC= 180°-(90°+45°)=45°

Remember that [tex]sin \alpha = \frac{opposite leg}{hypotenuse} [/tex], so [tex]oppositeleg=(hypotenuse)(sin \alpha )[/tex]

We know that hypotenuse= BC= 3in and [tex] \alpha [/tex]=∠DBC)=45°, so replacing the values we get:
[tex]CD=3sin(45)=2.12[/tex]

We can conclude that the segment CD is 2.12 in