as a hot air balloon rises vertically, its angle of elevation from point P on level ground 110 kilometers from th epoint Q directly underneath the balloon changes from 15 degree 5' to 31 degree 45'.

Approximately how far does the balloon rise during this period?
a. 38.4 km
b. 97.7 km
c.106.5 km
d. 86.5 km
e. 199.7 km
i need help solving this

If the balloon rises from height h to height H,

h/110 = tan(15°5')
H/110 = tan(31°45')

Distance risen = H-h

To solve this problem, we can use trigonometry and the concept of angle of elevation.

First, let's convert the angle measurements to decimal degrees.

15 degrees 5 minutes = 15 + 5/60 = 15.083 degrees
31 degrees 45 minutes = 31 + 45/60 = 31.75 degrees

Now, let's draw a diagram to visualize the situation:

P Q
|----------------|------- Balloon
110 km

We can consider the triangle formed by the balloon, point P, and point Q.

Let's call the distance the balloon rises h (in km).

From the diagram, we can see that the tangent of the angle of elevation is given by:

tan(31.75 degrees) = h / 110 km

Rearranging the equation, we can solve for h:

h = 110 km * tan(31.75 degrees)

Using a calculator, we find that tan(31.75 degrees) ≈ 0.6457.

h ≈ 110 km * 0.6457
h ≈ 71.027 km

Therefore, the balloon rises approximately 71.027 km during this period.

The closest option is c. 106.5 km.

So, the answer is c. 106.5 km.

To solve this problem, we can use trigonometry and the concept of similar triangles.

First, let's draw a diagram to represent the situation. We have a hot air balloon directly above point Q, with point P on the ground 110 kilometers away from Q.

```
P
|\
| \
| \
| \
| \
| \
110 | \ h
| \
| \
| \
| \
| \
Q-----------B
```

Let's consider the right-angled triangle BQP, where angle BPQ is the angle of elevation.

Given that the angle of elevation changes from 15 degrees 5 minutes to 31 degrees 45 minutes, we can find the difference in the angles:

Angle difference = (31° 45' - 15° 5') = 16° 40'

Now, let's label the sides of the triangle:

- The side opposite angle BPQ is the height of the balloon, which we'll call 'h' (what we are trying to find).
- The side adjacent to angle BPQ is the horizontal distance from point Q to point P, which is 110 kilometers.

Using trigonometry, we can write:

tan(angle BPQ) = height/horizontal distance

tan(angle BPQ) = h/110

tan(angle BPQ) = h/110

To find 'h', we can rearrange the equation:

h = 110 * tan(angle BPQ)

Now, we need to convert the angle from degrees to radians before substituting it into the equation:

Angle BPQ in radians = (16° 40') * (π/180°) ≈ 0.290900

Substituting this value back into the equation, we get:

h = 110 * tan(0.290900) ≈ 36.031352

So, the balloon rises approximately 36.031 kilometers during this period.

Looking at the answer choices provided:

a. 38.4 km
b. 97.7 km
c. 106.5 km
d. 86.5 km
e. 199.7 km

The closest answer is 38.4 km (option a). However, the estimated value we calculated is around 36.031 km, which is not exactly the same as any of the given answer choices. It's possible that the calculated value is an approximation.