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 answer to the question above is letter C. To explain the answer if the given question, a circle of 30 inches radius, if the central angle is 35 degrees, intersecting the circle forms an arc of length which is 18.33 inches.
Answer:
105
Step-by-step explanation:
Theorem:
If parallel lines are cut by a transversal, then same-side interior angles are supplementary.
Lines AB and CD are parallel, and angles 3 and 6 are same-side interior angles, so by the theorem above, angles 3 and 6 are supplementary. That means that the sum of their measures is 180 deg.
m<3 + m<6 = 180
m<3 + 75 = 180
Subtract 75 from both sides.
m<3 = 105
Answer: 105 degrees
R = 7
t = 3
d = rt
d = (7)(3)
d = 21
