The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Sextant Theorem

Date: 05/13/2001 at 21:55:23
From: Yolanda 
Subject: Mathematical theorem behind using a sextant.

The sextant is in essence a machine used to measure the distance 
between a celestial body and the horizon, and can be used for varying 
the angle between two parallel mirrors by using the Principle of 
Double Reflectivity and the phenomenon that the angle of the departing 
light ray will have been changed by double the angle between the 
mirrors. For example, if a body is 70 degrees above the horizon, the 
parallel mirrors of the sextant must be displaced by 35 degrees in 
order to send its reflected light rays straight through the optics.

There is a mathematical theorem that proves this. What is it, and what 
is its proof?

Thank you.

Date: 05/15/2001 at 12:58:34
From: Doctor Rick
Subject: Re: Mathematical theorem behind using a sextant.

Hi, Yolanda.

The first thing you need is a figure. I made one using the Geometer's 
Sketchpad program:

The path of light from the celestial body to the eye is shown in red: 
ABCD. The two mirrors are shown in green. The celestial body is on the 
continuation of BA upward. The horizon is on the continuation of DC to 
the left. The bottom mirror is half-silvered, or silvered on one half, 
so the horizon can be seen through the mirror. The upper mirror is 
rotated until BA is in line with the star.

The normals (perpendicular lines) to the mirrors are BE and CF. If the 
two mirrors were parallel, then the normals would be parallel. The 
deviation from parallel is measured by the angle BGC between the 
normals. It is this angle that is measured by the sextant (though not 
at point G).

When light is reflected in a mirror, the "angle of incidence" equals 
the "angle of reflection." These angles are the angles that the 
entering and reflected rays make with the normal to the mirror.

The theorem you seek can now be stated: If angle ABE equals angle EBC, 
and angle BCF equals angle FCD, then angle BHC (between the line of 
sight to the celestial body and the line of sight to the horizon) is 
twice angle BGC (between the normals to the mirrors).

Once all this preparatory work has been done, the geometrical proof is 
simple. Angle EBC is an exterior angle of triangle BGC, and thus its 
measure is equal to the sum of the measures of angle BGC and angle 
BCG. Therefore

(1)  m<BGC = m<EBC - m<BCG

Turning our attention to triangle BCH, angle ABC is an exterior angle, 
and thus its measure is equal to the sum of the measures of angle BCH 
and angle BHC. Therefore

(2)  m<BHC = m<ABC - m<BCH


(3)  m<ABC = m<ABE + m<EBC
           = m<EBC + m<EBC
           = 2*m<EBC

(4)  m<BCH = m<BCF + m<FCH
           = m<BCF + m<BCF
           = 2*m<BCF
           = 2*m<BCG

Therefore, substituting (3) and (4) in equation (2), we have

(5)  m<BHC = 2*m<EBC - 2*m<BCG
           = 2*(m<EBC - m<BCG)
           = 2*m<BGC

using equation (1) for the last step. Quod erat demonstrandum (Latin 
for "which is what we intended to prove").

- Doctor Rick, The Math Forum   
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Physics/Chemistry
High School Practical Geometry

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.