Another Hugh Worthington Approximation for Right Triangles.

A while back I wrote about a nice hand calculation of the angles in a right triangle which appeared in an article, The Resolution of Triangles(1780), by Rev. Hugh Worthington. The article prompted some really nice math from several bloggers leading to an understanding that it was essentially a version of Taylor's method applied to the ArcTan function and converted to degrees. A good explanation is at the Theorem of the Day site.

So today I thought I would introduce another of Rev Worthington's hand calculation approaches for a right triangle, this one with the angles and one side given, to find another side. Worthington credits this method at a Mr. Henry Watson from Navigation new modelled (abt 1715). The Watson method is shortened into a shorter proportion, which I will give below, but I begin with Watson's method of using an "artificial number".

The method works two ways, to find a smaller angle if the Hypotenuse is known, and then to find the Hypotenuse if a leg is known.
I will illustrate first the method using a given leg.
First, find the compliment of the angle opposite the side given, and square it, then multiply the result by four. Now divide that amount by 300 + three times the compliment. Finally add that amount to the angle again. This is the "artificial number" which he says is sometimes called the natural radius (suggestions why?). I will call this artificial number N and the acute angle we use A.

Then \( \frac{N}{hyp} = \frac{A}{a} \)

Suppose we begin with a right triangle with a side a = 5, and A= 40^o, B=50^o and C is the right Angle.

We take the compliment of 40, and square it to get 2500. Four times this amount gives 10,000. Now we divide by 300+ 3 times the compliment, (450) to get 22 2/9 Finally we add back the 40 to get 62 2/9 for the artificial number.
So the proportion is \( \frac{62 2/9}{ Hyp }= \frac{40}{5} \) ,   giving us a Hypotenuse of 7 7/9 . My calculator gives Sin^-1 \( (\frac{5}{7\frac{7}{9}}) \) as 40.0052.

Which I'm calling pretty darn close.

To find a leg when the hypotenuse is the side given, we use the compliment of the angle opposite the side we seek to know, and proceed in similar fashion.   This time, for ease of computation, let's let the  Hypotenuse be 5, and keep the angles at 40 and 50.  We want to find the side opposite the forty degree angle, so 50 is still the compliment of the angle so we have the same artificial number. 

Our proportion for this looks like \(  \frac{62 2/9}{ 5 }= \frac{40}{a}\)   giving  a= 45/14 or 3.2142857....   Checking again, we get 5 * sin (40^o = 3.21938 .... pretty close again.


The Rev Worthington gives the shortcut (modified for the letters used above) to directly calculate without the intermediate of the artificial number.

\( \frac{57.3}{b} + \frac{3b}{1000}  = \frac{hyp}{a} \) 

I'm hoping that any high school student worth their salt can figure out the presence of 57.3, but if not, see the comments of the previous post linked above. 

So who was this mathematical minister, this preacher of trigonometric hand calculations? A note about his popularity as a preacher and a brief history is here, and includes a reference to this "Resolution of Triangles paper, and no others. Perhaps it was his sole excursion into math.