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In ΔABC, m∠ACB = 90°, CD ⊥ AB and m∠ACD = 45°. Find: CD, if BC = 3 in
5 months ago
Q:
In ΔABC, m∠ACB = 90°, CD ⊥ AB and m∠ACD = 45°. Find: CD, if BC = 3 in
Accepted Solution
A:
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