Askyscraper is 396 meters tall. at a certain time of day, it casts a shadow that is 332 meters long. at what angle is the sun above the horizon at that time? 57° 50° 40° 33°

B. .

Step-by-step explanation:

Let the sun be x degrees above the horizon.

We have been given that a skyscraper is 396 meters tall. At a certain time of day, it casts a shadow that is 332 meters long.

We can see from our attachment that skyscraper and shadow forms a right triangle with respect to ground.  

We know that tangent relates the opposite side of right triangle to its adjacent, so we can set an equation as:

Using arctan we will get:

Therefore, the sun is 50 degrees above the horizon at that time.

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!