Question

in a Billiards game, peter hits a ball that is 20 in. from the wall The ball travels 34 inches until it hits the wall and bounces to a position that is 16 inches from the wall what is the distance X the ball traveled after it bounces off the wall to get to the ending position

Answer 1

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