A pilot is flying over a straight highway. He determines the angles of depression to two mileposts, 5.1 km apart, to be 28° and 51°, as shown in the figure.
5.1 km
28°
51°
B
A
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Find the distance of the plane from point A.
distance from A =
km
Find the elevation of the plane.
height =
km

To find the distance of the plane from point A and the elevation of the plane, we can use trigonometry. Let \( h \) be the elevation of the plane, \( d \) be the distance from the plane to point A along the ground, and \( d' \) be the distance from the plane to point B along the ground. The problem states that the distance between A and B is 5.1 km.

We have two right triangles to consider:
1. The triangle with angle of depression \( 28^\circ \) to point A.
2. The triangle with angle of depression \( 51^\circ \) to point B.

Using these angles, we can set up the following equations using the tangent function, which relates the angle of depression to the elevation of the plane and the horizontal distance:

\[ \tan(28^\circ) = \frac{h}{d} \]
\[ \tan(51^\circ) = \frac{h}{d'} \]

We also know that:
\[ d' = d - 5.1 \text{ km} \]

We can solve for \( d \) and \( d' \) in terms of \( h \):

\[ d = \frac{h}{\tan(28^\circ)} \]
\[ d' = \frac{h}{\tan(51^\circ)} \]

Since \( d' = d - 5.1 \):

\[ \frac{h}{\tan(51^\circ)} = \frac{h}{\tan(28^\circ)} - 5.1 \]

We can solve for \( h \):

\[ \frac{h}{\tan(51^\circ)} = \frac{h}{\tan(28^\circ)} - 5.1 \]

Multiply through by \(\tan(51^\circ) \tan(28^\circ)\):

\[ h \tan(28^\circ) = h \tan(51^\circ) - 5.1 \tan(51^\circ) \tan(28^\circ) \]

\[ h (\tan(28^\circ) - \tan(51^\circ)) = -5.1 \tan(51^\circ) \tan(28^\circ) \]

\[ h = \frac{5.1 \tan(51^\circ) \tan(28^\circ)}{\tan(51^\circ) - \tan(28^\circ)} \]

Let's calculate the values of the tangents and then \( h \):

\[ \tan(28^\circ) \approx 0.5317 \]
\[ \tan(51^\circ) \approx 1.2349 \]

\[ h = \frac{5.1 \times 0.5317 \times 1.2349}{1.2349 - 0.5317} \]

\[ h = \frac{5.1 \times 0.6566}{0.7032} \]

\[ h \approx \frac{3.34966}{0.7032} \]

\[ h \approx 4.76 \text{ km} \]

Now, we can find \( d \):

\[ d = \frac{h}{\tan(28^\circ)} = \frac{4.76}{0.5317} \approx 8.95 \text{ km} \]

So, the distance from the plane to point A is approximately \( 8.95 \text{ km} \), and the elevation of the plane is approximately \( 4.76 \text{ km} \).

**Summary:**

- Distance from A: \( 8.95 \text{ km} \)
- Elevation of the plane: \( 4.76 \text{ km} \)

Would you like more details on this calculation, or do you have any questions? Here are some questions you might be interested in:

1. How can trigonometric functions be used to solve real-world problems?
2. What are some applications of angles of depression and elevation in aviation?
3. How do you determine which trigonometric function to use in a given problem?
4. Can you explain the difference between angle of depression and angle of elevation?
5. What are some common mistakes to avoid when solving problems involving angles of depression and elevation?

**Tip:** Always double-check your angle measurements and ensure your calculator is set to the correct mode (degrees or radians) for the problem at hand.

Learn how to calculate the distance of a plane from point A and its elevation using trigonometry. This problem involves angles of depression and requires application of the tangent function. Suitable for Grades 10-12 students studying trigonometry.

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