Question

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

Answer 1

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

Remember that $sin \alpha = \frac{opposite leg}{hypotenuse}$, so $oppositeleg=(hypotenuse)(sin \alpha )$

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

We can conclude that the segment CD is 2.12 in

Answer 2

Given ΔABC,

m∠ACB = 90 degrees,

CD ⊥ AB

m∠ACD = 45 degrees

BC = 3 in

CD is unknown

 

To know the Length of CD, we use the trigonometric function Cos Ɵ = A/H

If ∠BCD is Ɵ,

H = BC = 3 in

A = CD (Unknown)

 

m∠ACD + m∠BCD = m∠ACB

m∠BCD = m∠ACB – m∠ACD

m∠BCD = 90 degrees – 45 degrees

m∠BCD = 45 degrees

Therefore, Ɵ = m∠BCD = 45 degrees

 

Cos Ɵ = A/H

Cos 45 = CD/3

CD = 3 Cos 45

CD= 3 x 0.7071

CD = 2.1213

CD is approximately 2.12 in