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

the answer to this question is in the pic

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl