6. tan2 (45° – c) = tan † (4 − a) cot † (A + a).

The method of proof is similar to that given in the other cases.

Exercise 85. Formulas of Right Triangles

1. From the formula cos c = cos a cos b show that the hypotenuse of a right spherical triangle is less than 90° if the two sides are both less than 90° or are both greater than 90°.

2. As in Ex. 1, show that the hypotenuse is greater than 90° if one side is greater than 90° and the other side less than 90°.

3. From the formula cos A = cos a sin B show that in a right spherical triangle an oblique angle and the opposite side are either both greater than 90° or both less than 90°.

From the formulas on pages 193-195 state the inferences to be drawn respecting the values of the other parts when:

4. c = 90°. 5. a = 90°.

6. b = 90°.
7. c = a.

8. a = b.
9. A= 90°.

10. c 90° and a = 90°.
11. a =
90° and b = 90°.

176. Napier's Rules. The ten formulas given on page 191 were very ingeniously reduced to two simple rules by John Napier, the inventor of logarithms. Since the right angle does not enter into the formulas, only five parts need be considered. Napier found that he could greatly simplify the treatment by considering:

These parts are shown in the above triangle, C being omitted because it is not used. Since, as we shall see, it is convenient to consider any one of these as the middle part and the other parts as the adjacent parts and the opposite parts, they are often arranged on a circle as shown, and are known as circular parts.

If we speak of b as a middle part, Co-A and a are the adjacent parts and Co-c and Co-B are the opposite parts.

The rules are as follows:

1. The sine of any middle part is equal to the product of the tangents of the adjacent parts.

2. The sine of any middle part is equal to the product of the cosines of the opposite parts.

These rules are easily remembered by the expressions tan. ad. and cos. op. While it is possible to get along very well without these rules, using the formulas on page 191, this is a convenient way of memorizing them.

177. Napier's Rules Verified. The correctness of Napier's rules may be easily shown by taking in turn each of the five parts as the middle part, and comparing with the formulas on page 191.

For example, let Co-c be taken as the middle part; then Co-A and Co-B are the adjacent parts, and a and b the opposite parts, as is seen from the figure. Then, by Napier's rules,

or

and

or

=

sin (Co-c) tan (Co-A) tan (Co-B),

cos c = cot A cot B ;

sin (Co-c)= cos a cos b,

cos c = cos a cos b.

These results agree with formulas 10 and 1 on page 191.

Exercise 86. Spherical Triangles

Deduce eight of the formulas on page 191 by means of Napier's rules, taking for the middle part :

7. What do Napier's rules become if we take as the five parts of the triangle the hypotenuse, the two oblique angles, and the complements of the two sides?

8. Solve a spherical right triangle, given a, b, and c.

9. Solve a spherical right triangle, given A, B,

and c.

10. Solve a spherical right triangle, given A, a, and b.

Find the number of degrees in the sides of a spherical triangle, given the angles of its polar triangle as follows:

11. 82°, 77°, 69°.

12. 8440, 8130, 721°.

13. 78° 30', 89°, 102°.

14. 83° 40', 48° 57', 103° 43'.

15. 96° 37' 40", 82° 29' 30", 68° 47'.

16. 43° 29' 37", 98° 22′ 53′′, 87° 36′ 39′′.

Find the number of degrees in the angles of a spherical triangle,

given the sides of the polar triangle in Exs. 17-20 :

18. 78° 47' 29", 106° 36' 42", a quadrant.

19. 111° 29' 43", a quadrant, a quadrant.

20. A quadrant, half a quadrant, three fourths of a quadrant.

21. The angles of a spherical triangle are 70.5°, 80.7°, and 101.6°. Find the sides of the polar triangle.

22. The sides of a spherical triangle are 40.72°, 90°, and 127.83°. Find the angles of the polar triangle.

23. Show that, if a spherical triangle has three right angles, the sides of the triangle are quadrants.

24. Show that, if a spherical triangle has two right angles, the sides opposite these angles are quadrants, and the third angle is measured by the opposite side.

25. How can the sides of a spherical triangle, measured in degrees, be found in units of length, when the length of the radius of the sphere is known?

178. Solution of the Right Spherical Triangle. By using either Napier's rules or the formulas on page 191, we can solve any right triangle if two parts besides the right angle are given.

It is a little easier to use the formulas, but the student who prefers to remember only Napier's rules can get on easily without charging his memory with the formulas or referring to page 191. The formulas given in the following solutions are all found on page 193.

179. Given Two Sides. Given the two sides a and b of the right spherical triangle ACB, solve the triangle.

For example, in the right spherical triangle ACB, given a = 27°

28' 36", b = 51° 12' 8", solve the triangle.

If we know the diameter or the radius of the sphere, say in feet, we can find the circumference, and thus compute c in feet.

If c is very near 0° or 180°, it may be found to a greater degree of accuracy first by computing B from the formula tan B = tan b csc a, and then computing c from the formula tan c = tan a sec B.

21. How many degrees are there in the arc of a great circle drawn from a point on the equator in longitude 40° E. to a point on the prime meridian in latitude 40° N.?

22. Greenwich lies on the prime meridian 51° 28' 38" N. The arc of a great circle drawn from Greenwich to a point on the equator in longitude 25° W. makes what angle with the equator?

23. The arc of a great circle drawn from Greenwich to a point on the equator in longitude 150° E. makes what angle with the prime meridian?

24. How many degrees are there in the arc of a great circle drawn from a point on the equator in longitude 0° to a point in longitude 48° W., latitude 30° N.?

25. In a right spherical triangle on a sphere of radius 6 in. it is given that a = 45° and b = 70°. Find the length of e in inches.

*

26. In a right spherical triangle on a sphere of diameter 2 ft. it is given that a = ÷ 75° and b = 75°. Find the length of c in inches. 27. Taking the radius of the earth as 4000 mi., how many miles is it, on a great circle, from a point on the equator in longitude 70° W. to a point on the prime meridian in latitude 60° N. ?

28. The arc of a great circle drawn from a point on the prime meridian 60° N. to a point on the equator 60° W. makes what angle with the prime meridian and with the equator?

29. In Ex. 28, what is the length of the arc, taking the radius of the earth as 4000 mi.?