# EXERCISE 30 — LATITUDE BY MERIDIAN ALTITUDE SUN (Numerical Solution)

1. ###### ON 14th Sept 2008, in DR longitude 116˚ 27.0’ W, the sextant meridian altitude of the sun’s UL North of the observer was 70˚ 29.8’. If IE was 3.2’ off the arc and HE was 12m, find the latitude and state the direction of the PL (LOP).

• Prem says:

I wanna know, in question no. 1, the Dr longitude given is West, but why is the LIT in the calculation East

• Vikash Kumar says:

That’s a typing mistake otherwise the calculation is correct

• Captain America says:

In 3rd problem how to find true altitude position if lat is not given

How to find dec is greater than lat or dec smaller than lat

• Rizzqi says:

To find true altitude you don’t need latitude and to identify dec and lat which greater is look at mzd and dec poles. If mzd n and dec n then plus if differences poles minus. Answer will depend on either dec and mzd who have highers value.

• Angelbert Alban says:

Where can I find a free 2008 Nautical Almanac (NP314-08). Thanks.

• munir rancho says:

• Shubham says:

Please explain how to decide true altitude position N or S when DR lat is not given

• Chiranjay says:

True alt N or S is same as azimuth at mer pass

• Mohd khaleel says:

How can identify that true alt “s” or N”

• Abhishek Manral says:

it is named North or South, with respect to observer and the declination, if observer is north and declination is south then Cel body is south, so you can use this.

• Nitish Kumar Maurya says:

In the 5th solutions how the sextant altitude is 61°27.5′. I couldn’t understand

• Joel says:

same as bearing

• HASSAN SESAY says:

On the 21st jan.2008, the dip(HE10m) 05.6′ is subtracted from the observed Alt. where is the 05.6′ to arrived at the App Alt and 16.1′ Tcorrn.LL added to App Alt, to arrive at TAlt, MZD 04 44.4 S, not in the solution and Dec 19 54.4 s also not in the solution and finally you arrived at an answer
latitude 24 38. 8′ where are all these figures from