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 :
\(sin\theta=\frac{perpendicular}{hypotenuse}\)
IN ΔABC perpendicular = AB=20 inches and hypotenuse = AC=34 inches
\(sinC=\frac{AB}{AC}\)
\(sinC=\frac{20}{34}\)
IN ΔEDC perpendicular = ED=16 inches and hypotenuse = EC=x inches
\(sinC=\frac{ED}{EC}\)
\(sinC=\frac{16}{x}\)
Since ∠ACB = ∠ECD
⇒\(\frac{16}{x}=\frac{20}{34}\)
⇒\(\frac{16*34}{20}=x\)
⇒\(27.2=x\)
Thus distance X the ball traveled after it bounces off the wall to get to the ending position =27.2 inches
