The answer to this question is c i think
The horizontal distance is given by the formula:
tan θ=opposite/adjacent
where:
θ=21
opposite=2000 ft
adjacent=a
thus calculating for a we get:
tan 21=2000/a
thus
a=2000/tan 21
a=5210.18 ft
Answer: 5210.18 ft
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".
That would be 31ft 6ins x 54ft.
Live long and prosper.
the sum of 12 and a number/n
