You need to use a ratio of height (H) to shadow length (L) to solve the first problem. It's basically a use of similar triangles, with two perpendicular sides, and with the shadow making the same angle with the vertical.
6 ft = 72 ins, so that rH/L = 72/16 = 9/2 for the player.
So the bleachers are 9/2 x 6 ft = 27 ft.
For the second problem, 9 ft = 108 in, so that the ratio of the actual linear dimensions to the plan's linear dimensions are 9ft/(1/2in) = 2 x 108 = 216.
So the stage will have dimensions 216 times larger than 1.75" by 3".
B. a pair of alternate exterior angles with measures of 130° and 50°
Step-by-step explanation:
Alternate exterior angles formed when two parallel lines are cut across by the same transversal line, are said to be congruent.
Therefore, a pair of alternate exterior angles cannot have different angle measures of 130° and 50°. Rather, both alternate exterior angles should be of equal measure of degree.
All other scenarios are possible result except option B: "a pair of alternate exterior angles with measures of 130° and 50°".
