Mi Lighthouses
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math help?
could you explain to how to do this as well as the answer.
1. The angle of elevation from a boat to the top of a lighthouse is 35°. The lighthouse is 96 ft tall. How far from the base of the lighthouse is the boat?
a 7 ft
b137 ft
c 167 ft
d 6 ft
2. Ming launched a model rocket from 20 m away. The rocket traveled straight up. Ming saw it peak at an angle of 70°. If she is 1.5 m tall, how high did the rocket fly?
a 55 m
b 2 m
c 20 m
d 57 m
3. An airplane is flying 2.5 mi above the ground. If the pilot must begin a 3° descent to an airport runway at that altitude, how far is the airplane from the beginning of the runway (in ground distance)?
a 48 mi
b 47 mi
c 5 mi
d 8 mi
For these trigonometry questions, remember the following line:
SOH CAH TOA (Suck a Toe?)
This means, in a right angle triangle,
Sin(x) = Opposite / Hypotenuse
Cos(x) = Adjacent / Hypoetenus
Tan(x) = Opposite / Adjacent
So, let's solve these questions.
1. The angle of elevation from a boat to the top of a lighthouse is 35°. The lighthouse is 96 ft tall. How far from the base of the lighthouse is the boat?
When you draw this diagram out, you see instantly that it is a right-angle triangle. This allows us to use the above equations. Since we have the angle (x), as well as the opposite (96), and we need to find the line that is adjacent, the best equation to use in this case would be tan(x).
tan(x) = Opposite / Adjacent
tan(35) = 96 / Adjacent
Adjacent = 96 / tan(35)
Adjacent = 137.142
So, the lighthouse is approximately 137 ft away from the boat. So the answer is b. 137 ft
2. Ming launched a model rocket from 20 m away. The rocket traveled straight up. Ming saw it peak at an angle of 70°. If she is 1.5 m tall, how high did the rocket fly?
So, after drawing out this diagram, we see that Ming is 20m away from the right-angle, and the angle of elevation from where her EYE LEVEL is, is 70°. We need to find out how high the rocket went. So, since we have the angle, as well as the adjacent line, and we need to find the opposite line, the best equation to use would be one again, tan(x).
tan(x) = Opposite / Adjacent
tan(70) = Opposite / 20
Opposite = tan(70) * 20
Opposite = 54.95
But since she is 1.5m tall, we need to add 1.5 to the height, since she "saw" it peak at an angle of 70°. So, the height the rocket flew was 55 + 1.5 = 56.5, which is approximately 57m. So, the answer should be d. 57 m
3. An airplane is flying 2.5 mi above the ground. If the pilot must begin a 3° descent to an airport runway at that altitude, how far is the airplane from the beginning of the runway (in ground distance)?
After drawing out this diagram, we can see that we have the angle of 3° on the OUTSIDE of the right angle triangle. So, we need to find the degrees on the inside. Since the outside of that angle is 3°, then that means the other angle within the right-angle triangle is also 3° (the angle from where the plan will land). We also know the height of 2.5mi of the plane. We need to find the ground distance, or line that is adjacent to the 3°. So, we have the degrees (x), the adjacent, and the opposite(2.5). The best equation to use, once again, would be tan(x).
tan(x) = Opposite / Adjacent
tan(3) = 2.5 / Adjacent
Adjacent = 2.5 / tan(3)
Adjacent = 47.7
So, the airplane is approximately 48 miles from the beginning of the runway in ground distance, so the answer is:
a. 48 mi
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