<p>This is basically a trigonometry question.  I wasn't sure from the question if the apparent location from your friend's position was six degrees higher or lower.  I chose higher for the example, but you could change it.<br>
<img src="https://i.sstatic.net/ysmtn.jpg" alt="meteor altitude"></p>

<p>You have two triangles with the same altitude, but different bases and angles.</p>

<p>$$\text{altitude} = x \tan(81^{\circ}) = (x + 20km) \tan (75^{\circ})$$  Do you see how to calculate it from there?</p>

CC BY-SA 3.0
2015-02-04T05:33:51.823Z
1: You are making the down range assumption, which is what I think the question intends, but does not state. 2: As this is an astro SE and not a trig homework site I would mention just doing a scale diagram as a viable method.  3: As astro is closer to physics than trig I would give an approximate solution method using small angles. 4: Also you may feel the need to justify the flat Earth approximation (which is OK but you need to think about these things)
Conrad Turner
CC BY-SA 3.0
2015-02-04T01:25:37.513Z
2015-02-04T01:25:37.513Z
BowlOfRed
<blockquote>
  <p>One night, you and your friend were talking to each other. Your friend far about 20 km. from you. You told your friend that you see the shooting star pass through the sky at altitude of 75 degrees. Your friend also see this shooting star, but has difference apparent location by 6 degrees. Find the location of this shooting star, what is it hight from earth ground.</p>
</blockquote>

<p>I think 6 degrees is the different of azimuth angle. But as I live in different lattitude, it may be effect the distance between friend and me. Or the answer is just 75degrees turn to the height from the earth ground. But how? Which and how does the correct way to solve this question?</p>

CC BY-SA 3.0
2015-02-02T17:48:48.710Z
What do you want to know, how to calculate it or just what it is (approximatly)? (its ~170-180km ish assuming your friend is directly up or down range)
Conrad Turner
CC BY-SA 3.0
2015-02-03T23:28:46.973Z
I want to know how to calculate it. :)
user9686
<star>
Find the height of the shooting star from the Earth's ground
user9686
# Find the height of the shooting star from the Earth's ground

(Asked by: username_0 on 2015-02-02)

> One night, you and your friend were talking to each other. Your friend far about 20 km. from you. You told your friend that you see the shooting star pass through the sky at altitude of 75 degrees. Your friend also see this shooting star, but has difference apparent location by 6 degrees. Find the location of this shooting star, what is it hight from earth ground.

I think 6 degrees is the different of azimuth angle. But as I live in different lattitude, it may be effect the distance between friend and me. Or the answer is just 75degrees turn to the height from the earth ground. But how? Which and how does the correct way to solve this question?

**Tags:** star
> **Comments:**
> **username_1**: What do you want to know, how to calculate it or just what it is (approximatly)? (its ~170-180km ish assuming your friend is directly up or down range)  
> **username_0**: I want to know how to calculate it. :)  

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## 1 answers

### Answer by username_2 on 2015-02-04. Score: 4 | Accepted answer

This is basically a trigonometry question. I wasn't sure from the question if the apparent location from your friend's position was six degrees higher or lower. I chose higher for the example, but you could change it.  
![meteor altitude](https://i.sstatic.net/ysmtn.jpg)

You have two triangles with the same altitude, but different bases and angles.

$$\text{altitude} = x \tan(81^{\circ}) = (x + 20km) \tan (75^{\circ})$$ Do you see how to calculate it from there?

> **Comments:**
> **username_1**: 1: You are making the down range assumption, which is what I think the question intends, but does not state. 2: As this is an astro SE and not a trig homework site I would mention just doing a scale diagram as a viable method.  3: As astro is closer to physics than trig I would give an approximate solution method using small angles. 4: Also you may feel the need to justify the flat Earth approximation (which is OK but you need to think about these things)  

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