*Approximate the angle of elevation of the sun if a box 5 meters tall casts a shadow 4 meters long on level ground. *As a weather ballon rises vertically its angle of elevation from a point P on the level ground 110km.from the point Q directly underneath the ballon changes from 19degrees 20 minutes to 31 degrees 50 minutes.Approximate how far does the ballon rises during the period.

No spherical trig here. The balloon problem might have qualified, but it says "level ground."

(a) tanθ = 5/4
(b) review your tangent function. If the balloon travels a distance d, then

d = 110 tan31°50' - 110 tan19°20'

To approximate the angle of elevation of the sun, you can use the concept of similar triangles.

In this scenario, the box is the vertical object, and its shadow acts as the horizontal object. The height of the box is 5 meters, and the length of its shadow is 4 meters. Let's call the angle of elevation θ.

By considering the two similar triangles, we can write the following proportion:

tan(θ) = height of the box / length of the shadow
tan(θ) = 5 / 4

Now, we can solve for θ by taking the inverse tangent (arctan) of both sides:

θ ≈ arctan(5/4)

Using a calculator, we can find the arctan(5/4) to be approximately 51.34 degrees.

Therefore, the approximate angle of elevation of the sun is 51.34 degrees.

Now let's move on to the second question about the weather balloon.

To approximate how far the balloon rises during the given angle change from 19 degrees 20 minutes to 31 degrees 50 minutes, we can use the concept of trigonometry and the difference of the angles.

Let's convert the angles to decimal form for calculations:

19 degrees 20 minutes = 19.33 degrees
31 degrees 50 minutes = 31.83 degrees

The change in angle is 31.83 - 19.33 = 12.5 degrees.

Since the motion is vertical, we can consider the right triangle formed by the motion of the weather balloon. The vertical distance the balloon rises can be represented by the opposite side of the triangle, and the distance between point P and Q can be considered as the adjacent side of the triangle.

By using the tangent function:

tan(change in angle) = opposite side / adjacent side

tan(12.5 degrees) = opposite side / 110 km

Now, we can solve for the opposite side (vertical distance):

opposite side = tan(12.5 degrees) * 110 km

Using a calculator, we find that the opposite side is approximately 23.5 km.

Therefore, the approximate distance that the balloon rises during the period is 23.5 km.