Answer:
27.2 inches
Step-by-step explanation:
Given :
Refer the attached figure
ΔABC AND ΔEDC are right angled triangle at B and D respectively
So we will use trigonometric ratios :
[tex]sin\theta=\frac{perpendicular}{hypotenuse}[/tex]
IN ΔABC perpendicular = AB=20 inches and hypotenuse = AC=34 inches
[tex]sinC=\frac{AB}{AC}[/tex]
[tex]sinC=\frac{20}{34}[/tex]
IN ΔEDC perpendicular = ED=16 inches and hypotenuse = EC=x inches
[tex]sinC=\frac{ED}{EC}[/tex]
[tex]sinC=\frac{16}{x}[/tex]
Since ∠ACB = ∠ECD
⇒[tex]\frac{16}{x}=\frac{20}{34}[/tex]
⇒[tex]\frac{16*34}{20}=x[/tex]
⇒[tex]27.2=x[/tex]
Thus distance X the ball traveled after it bounces off the wall to get to the ending position =27.2 inches
