<p>It is a strangely worded question I agree.</p>

<p>I would start with the sun appearing directly overhead at the latitude of the tropic of cancer 23.5 degrees. As you go north, vertical shifts north and the Sun appears to be lower towards the horizon due south. To make the Sun appear to be at 73 Degrees from horizontal you would need to be 17 degrees north of the tropic of cancer. 23.5+17=40.5</p>

<p>Local differences in gravitational direction and the eccentricities of the circumference of the Earth due to the equatorial bulge make the answer a bit uncertain hence the range </p>

<p>30 to 45 degrees.</p>

CC BY-SA 4.0
2019-12-08T18:18:57.293Z
Couldn't you be 17 degrees south of the tropic of cancer? That seems to be what the question is about.
ProfRob
CC BY-SA 4.0
2019-12-08T07:52:48.920Z
2019-12-08T18:14:58.147Z
2019-12-08T18:14:58.147Z
JohnRobyClayton
<p>From Question #6 of <a href="https://usaaao.org/resources/past-exams/" rel="nofollow noreferrer">USAAAO First Round 2018</a></p>

<blockquote>
  <ol start="6">
  <li>(1 point) For the following problem find the range in which the answer lies: on the day of summer solstice, on which geographic latitude is the sun culminating at the angle of +72° 50' north of the equator?
  
  <ul>
  <li>A. 0° − 15°</li>
  <li>B. 15° − 30°</li>
  <li>C. 30° − 45°</li>
  <li>D. 45° − 60°</li>
  <li>E. 60° − 75°</li>
  </ul></li>
  </ol>
</blockquote>

<p>I have two answers: 41° and 6°, but only one is accepted. No solutions are provided for 2018 and it is an objective question.</p>

CC BY-SA 4.0
2019-12-03T16:58:55.933Z
I find the question to be worded awkwardly. I wonder if "north of the equator" should have been written "north of the observer". In other words, the Sun could culminate at 73 degrees (above the horizon) if the observer is at 41 latitude (Sun culminates to the south) or if the observer is at 6 latitude (Sun culminates to the north). I suggest choosing 6 degrees N latitude.
JohnHoltz
<the-sun><earth><coordinate><positional-astronomy>
Latitude of given altitude of Sun
Chooi Je Qin
# Latitude of given altitude of Sun

(Asked by: username_0 on 2019-12-03)

From Question #6 of [USAAAO First Round 2018](https://usaaao.org/resources/past-exams/)

> 6. (1 point) For the following problem find the range in which the answer lies: on the day of summer solstice, on which geographic latitude is the sun culminating at the angle of +72° 50' north of the equator?
>    * A. 0° − 15°
>    * B. 15° − 30°
>    * C. 30° − 45°
>    * D. 45° − 60°
>    * E. 60° − 75°

I have two answers: 41° and 6°, but only one is accepted. No solutions are provided for 2018 and it is an objective question.

**Tags:** the-sun, earth, coordinate, positional-astronomy

> **Comments:**
> **username_1**: I find the question to be worded awkwardly. I wonder if "north of the equator" should have been written "north of the observer". In other words, the Sun could culminate at 73 degrees (above the horizon) if the observer is at 41 latitude (Sun culminates to the south) or if the observer is at 6 latitude (Sun culminates to the north). I suggest choosing 6 degrees N latitude.  

---

## 1 answers

### Answer by username_2 on 2019-12-08. Score: 1

It is a strangely worded question I agree.

I would start with the sun appearing directly overhead at the latitude of the tropic of cancer 23.5 degrees. As you go north, vertical shifts north and the Sun appears to be lower towards the horizon due south. To make the Sun appear to be at 73 Degrees from horizontal you would need to be 17 degrees north of the tropic of cancer. 23.5+17=40.5

Local differences in gravitational direction and the eccentricities of the circumference of the Earth due to the equatorial bulge make the answer a bit uncertain hence the range

30 to 45 degrees.

> **Comments:**
> **username_3**: Couldn't you be 17 degrees south of the tropic of cancer? That seems to be what the question is about.  

---

