Turner's first example of solving a right triangle.

Turner explains how his axioms of right triangles can also solve oblique triangles.

Turner's third example of solving an obtuse or oblique triangle.

     Reverend Turner continues his textbook by giving examples of triangles that have three known parts and need to be solved.  Each example requires applying the four theorems in a different order so the reader can learn in what case to use each of the theorems.  Not all four of the theorems are needed to solve every triangle but each of the four is used in a particular case.

     The first example, featured here, is Case I out of the four cases given by Turner for right triangles [1].  In this instance, the angles of the triangle are known, but only one of the lengths of the sides is given.  Since he shows all of his work in steps, we can follow along as Turner solves this case.  He begins by using his first axiom to solve for the hypotenuse of the triangle.  Since the first axiom has two parts, it is broken into two steps in the example.  For reference, here is the first step in notation:

     Turner squares the angle's complement, which in this case is angle A at 35.7 degrees.  Then he multiplies this answer by 4.  To the right of that column, Turner calculates the base of the fraction by multiplying 35.7 by 3 and then adding 300.  Now he has 5097.96, the numerator in the fraction, and long divides this out by the denominator, 407.1.  This answer, 12.5, is then added to angle C to get the Natural Radius, 66.8.

     With the Natural Radius acquired, the length of the triangle's hypotenuse can be found through a ratio.  The Natural Radius (66.8) is proportional to the length of the hypotenuse (x) just as the angle C (54.3) is proportional to its opposite side (78).  Therefore, Turner is able to set up the ratio as follows, and solve:

     Turner finishes solving this triangle by using his third axiom to find the remaining side length.  This requires finding the sum and the difference of the hypotenuse and the known side, multiplying them by each other, and then taking the square root. 

     In solving this triangle, Turner has walked the reader through using his axioms, and it did not require any sine or cosine functions.  By showing all of his work, the reader can clearly see that his formulas work as intended.  However, checking this with trigonometric functions and a modern calculator produces slight discrepancies.  For example, Turner's answer for the hypotenuse is 95.9 units, whereas in using a cosine function with a calculator the answer produced is 96.049 units.  This discrepancy could produce disastrous results in finer modern applications, but during Turner's day it was most likely not a cause for concern.

     A note: It seems that Turner's definition of the triangle at the top of the page as having angles A and C at 35.41 and 54.19 degrees is a mistake, for in his steps he appears to use instead 35.7 and 54.3 degrees, and 35.41 and 54.19 do not add together to 90 degrees as required by the rules of triangles.

Oblique Triangles

     So far, Turner has covered solving right triangles, but on page 10, seen here, he explains how his axioms of right triangles can also be used to solve oblique triangles [2].  Doing this requires dividing up the oblique triangle into right triangles and then solving each resultant right triangle.  Turner demonstrates this in his next four cases on the following pages.  Seen here is Case III, on page 13, in which the reader can follow along as Turner breaks down this triangle and finds its missing parts [3].