A billiard ball is struck by a cue. It travels 100 cm before ricocheting off a rail and traveling another 100 cm into a corner pocket. The angle between the path as the ball approaches the rail and the path after it strikes the rail is 45 degrees. How far is the corner pocket from where the cue initially struck the ball?

76.53 cm

Step-by-step explanation:

From the question, the triangle formed has 2 sides and an included angle ;

Angle, θ = C = 45°

We use cosine rule :

Taking the side to find as c

a = 100 ; b = 100

From cosine rule :

c = sqrt(a² + b² - 2abCosC)

c = sqrt(100² + 100² - 2(100*100)Cos45

c = sqrt(10000 + 10000 - 20000 * 0.7071067)

c = sqrt(20000 - 14142.135)

c = sqrt(5857.865)

c = 76.536690

c = 76.53 cm

step-by-step explanation:

we are given

since, we have to isolate x on anyone side

so, we can move all x terms on one side

subtract both sides by 2x

add both sides by 9

divide both sides by 4

we get

check above link.

graphically?   well, you graph them both, the solutions is where they both intersect each other, since there are two solutions, meaning they will meet twice, or namely the linear will run into the quadratic curve twice, and where they meet, those x,y coordinates are the solutions to the system.