A news helicopter 60 meters above the road spotted two accidents. The angles of depression are 10.2° and 8.7°. How are apart are the accidents? (The accidents are in the same general direction from the helicopter.) please help. i'm not sure what i am doing wrong!

the ground and the height of the helicopter form two sides of a right triangle

a line from the copter to the accident forms the hypotenuse

the angle between the vertical and the hypotenuse is ... (90º - 10.2º) for one accident
... and (90º - 8.7º) for the other

the distance to the farther accident is
... 60 m * tan(90º - 8.7º)

to the nearer one is
... 60 m * tan(90º - 10.2º)

To find out the distance between the two accidents, we can use the concept of trigonometry and the tangent function.

Let's assume the distance between the helicopter and the first accident is 'x' and the distance between the helicopter and the second accident is 'y'. Since the accidents are in the same general direction from the helicopter, both distances will be parallel to the road.

From the information given, we know that the angle of depression for the first accident is 10.2°, and the angle of depression for the second accident is 8.7°.

Using the tangent function, we can relate the angles of depression to the distances:

For the first accident:
tan(10.2°) = x/60 (1)

For the second accident:
tan(8.7°) = y/60 (2)

To solve for x and y, we need to rearrange equations (1) and (2):

x = 60 * tan(10.2°)

y = 60 * tan(8.7°)

Now, we can calculate the values:

x ≈ 10.370 meters

y ≈ 9.348 meters

To find the distance between the two accidents, we can subtract the smaller distance from the larger distance:

distance between the accidents = |x - y|

distance between the accidents ≈ |10.370 - 9.348|

distance between the accidents ≈ 1.02 meters

Therefore, the two accidents are approximately 1.02 meters apart.