Question

An observer (O) spots a plane flying at a 35° angle to his horizontal line of sight. If the plane is flying at an altitude of 17,000 ft., what is the distance (x) from the plane (P) to the observer (O)?

Answer 1

Do you have an image that goes with it??

Answer 2

http://hillsborough.flvs.net/webdav/assessment_images/educator_geometry/v15/module07/0704_g1_q2.jpg

Answer 3

thats instructions

Answer 4

It should have been the image.
Don't know why the link isn't working. I'm about as tech savvy as I'm good with math, though.

Answer 5

Got it?

Answer 6

That is it exactly. But the problem is, the lesson made no sense so I have no idea whatsoever on how to do the mathematical operations to figure it out.

Answer 7

You want me to keep going? or not?

Answer 8

I'd love it if you kept going and explained how this whole thing works.

Answer 9

To add on to my diagram, we are actually trying to find the hypotenuse of this triangle so we can use the angle we are given, and the altitude we are given to set up the sine of 35 degrees.

sin35 = opposite(altitude)/hypotenuse(x)
sin35 = 17,000/hypotenuse
17,000 time (sin35) = hypotenuse(x)
So now we use our calculator to fin sin35 and multiply it by 17,000 and then you have x!

Answer 10

Okay, I did that, but now I have -0.984807753. I feel like I did something incredibly wrong here.

Answer 11

hmmm wait a sec... shoot

Answer 12

Okay.

Answer 13

sorry messed up. give me a sec:)

Answer 14

Alrighty.

Answer 15

\[\sin(observer's \angle)=\frac{ altitude }{ x }----->x=\frac{ altitude }{ \sin(observer's \angle) }\]

Answer 16

Notice the picture I drew that should not be your answer. try this, 17,000 divided by the sin35

Answer 17

@NoahRS : Have you set your calculator to "Degree Mode?" If not, please do so.

Answer 18

Try finding sin 30 degrees. If you don't get 0.5 as the answer, you're in the wrong calculator mode.

Answer 19

It says DEG in the top corner, but it all just shows up in decimals.

Answer 20

An observer (O) spots a plane flying at a 35° angle to his horizontal line of sight. If the plane is flying at an altitude of 17,000 ft., what is the distance (x) from the plane (P) to the observer (O)?

Looking at the great diagram that Yana has drawn, identify the "opposite side" of the triangle as the height, 17000 feet. Note that this side is "opposite" the 35-degree angle. We want to find the hypotenuse of this triangle, which is the distance x from plane to observer.

Which trig function involves the angle, the opposite side and the hypotenuse?

Answer 21

Sine, which I've been trying to use.

Answer 22

Right. Then,\[\sin 35=\frac{ opp~side }{ hyp }=\frac{ 17000 }{ hyp }\]

Answer 23

Think about how you would solve this for "hyp."

Answer 24

Multiply thirty-five against 17000?

Answer 25

\[Hint:~ if~a=\frac{ b }{ c }, ~then~c=\frac{ a }{ b }\]

Answer 26

Are you telling me to use the pythagorean theorem? With the a2 + b2 = c2?

Answer 27

So, if \[\sin 35=\frac{ 17000 }{ hyp }, hyp=??\]

Answer 28

"Are you telling me to use the pythagorean theorem? With the a2 + b2 = c2?" No, not at all. Use trig here.

Answer 29

I don't understand how to get the hypotenuse at all. I'm assuming a is the altitude and b would be the base, but there is no base and there isn't any hypotenuse. I don't know what I'm doing here, that is why I came to this website. For a step-by-step walk-through of it for someone who can't grasp the basics of geometry or trigonometry.

Answer 30

\[\sin 35 = \frac{ 17000 }{ hyp }\] can be re-written as\[\frac{ \sin 35 }{ 1 }=\frac{ 17000 }{ hyp }\] and we can flip both fractions over to obtain\[\frac{ 1 }{ \sin 35 }=\frac{ hyp }{ 17000 }\] I'm now asking you to multiply both sides of this equation by 17000.