Answer:

Step-by-step explanation:
Think of a right triangle. It may help you to draw it.
Start at a point. Draw a point and label it point A. That is where the plane takes off from. Now draw a diagonal segment tilted up to the right and stop at a point and label it B. Segment AB is the path of the airplane. Now draw a segment down vertically to a point you will label C, so that points A and C are on the same horizontal line. Connect points A and C.
You now have a right triangle. Angle C is the right angle. AB is the hypotenuse and is 5.4 km. That is the actual distance the plane traveled. BC is 1.12 km (since 1120 m = 1.12 km) and is a leg of the right triangle. The angle you are looking for is angle A.
For angle A, BC is the opposite leg. AB is the hypotenuse.
The trig ratio that relates the opposite leg to the hypotenuse is the sine.





Answer: 12.0 degrees
9514 1404 393
Answer:
(-98, 18)
Step-by-step explanation:
We recognize the equation as being in the "point-slope" form.
y -k = m(x -h) . . . . . . . a line with slope m through point (h, k)
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Comparing this form to the given equation, we identify the parameters as ...
k = 18, m = 71.3, h = -98
The point on this line is (h, k) = (-98, 18).
Apparent magnitude isn’t the actual brightness, it’s what the brightness appears to be from earth, absolute brightness is the actual brightness an example is our sun it appears bright on earth but in space it is so much more brighter
The slope intercept form is -9,-4
I hoped this helped :)