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a search and rescue pilot is flying over a section of forest. The angle of depression to two clearings is 32 degrees and 56 degrees respectively. According to her co-pilot, the two clearings are 5.7km apart. find the distance from airplane to clearing A, rounded to nearest hundredth of a kilometre

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Let's call the position of the plane "P", and the positions of clearing A and clearing B "A" and "B", respectively. We want to find the distance from the plane to clearing A.

First, draw a diagram. The angle of depression to clearing A means that the line from the plane to clearing A makes a 32 degree angle with the horizontal. Similarly, the angle of depression to clearing B means that the line from the plane to clearing B makes a 56 degree angle with the horizontal. We know that the distance between clearings A and B is 5.7 km.

Now, we can use trigonometry to find the distance from the plane to clearing A. Let's call this distance "d". We can set up the following equation:

tan(32) = d/x

where x is the horizontal distance from the plane to clearing A. We can rearrange this equation to solve for d:

d = x tan(32)

Similarly, we can set up an equation for the distance from the plane to clearing B:

tan(56) = (x + 5.7) / d

We can rearrange this equation to solve for d:

d = (x + 5.7) / tan(56)

Now we can set the two expressions for d equal to each other and solve for x:

x tan(32) = (x + 5.7) / tan(56)

x tan(32) tan(56) = x + 5.7

x (0.6561) = x + 5.7

0.6561 x = x + 5.7

-0.3439 x = 5.7

x = -5.7 / 0.3439

x ≈ -16.57

Since the distance from the plane to clearing A is a positive distance, we can ignore the negative solution. Therefore, the distance from the plane to clearing A is approximately:

d = x tan(32) ≈ (-16.57) tan(32) ≈ 9.18 km

Rounding this to the nearest hundredth of a kilometer gives:

d ≈ 9.18 km

Therefore, the distance from the plane to clearing A is approximately 9.18 km.

:)?

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