B

BED:

Also, let a plane perpendicular to OA be passed through B, cutting OA at E and OC at D. Draw BE, BD, and DE. BE and DE are each to OA (Geom. § 462); therefore A. The plane BDE is to the plane AOC (Geom. § 518); hence BD, which is the intersection of the planes. BDE and BOC, is to the plane AOC (Geom. § 520), therefore to OC and

E

C

If in [38] we substitute for cos a and cos b their values from [41], we obtain

cos ccot A cot B.

1

[43]

NOTE. In order to deduce the second formulas in [39]-[42] geometrically, the auxiliary plane must be passed through A1 to OB.

These ten formulas are sufficient for the solution of any right spherical triangle.

In deducing these formulas, it has been assumed that all the parts of the triangle, except the right angle, are less than 90°. But the formulas also hold true when this hypothesis is not fulfilled.

Let one of the legs a be greater than 90°, and construct a figure for this case (Fig. 39) in the same manner as Fig. 38.

The auxiliary plane BDE will now cut both CO and AO produced beyond the centre O; and we have

Likewise, the other formulas, [39]–[43], hold true in this case. Again, suppose that both the legs a and b are greater than 90°. In this case the plane BDE (Fig. 40) will cut CO produced beyond 0, and 40 between A and O; and we have

cos COE = OD cos DOE

=(-cos a) (-cos b)

=cos a cos b,

a result agreeing with [38].

And the remaining formulas

may be easily shown to hold true.

Like results follow in all cases; in other words, Formulas [38]-[43] are universally true.

EXERCISE XXX.

1. Prove, by aid of Formula [38], that the hypotenuse of a right spherical triangle is less than or greater than 90°, according as the two legs are alike or unlike in kind.

2. Prove, by aid of Formula [41], that in a right spherical triangle each leg and the opposite angle are always alike in kind.

3. What inferences may be drawn from Formulas [38]-[43] respecting the values of the other parts: (i.) if c=90°; (ii.) if a=90°; (iii.) if c=90° and a=90°; (iv.) if a = 90° and b 90°?

Deduce from [38]-[43] and [18]-[23] the following formulas :

4. tan2=tan (c − a) tan 1⁄2 (c+a).

HINT. Use Formula [18] and substitute in it the value of cosb in [38].

5. tan2 (45°-4)=tan (c-a) cot (c+a). A) 1⁄2 †

8. tan2a=tan [1⁄2 (A + B) — 45°] tan [1⁄2 (A — B) +45°].

9. tan2 (45° —c) = tan † (A — a) cot1⁄2 (4+a).

10. tan2 (45° — 6) = sin (A — a) csc (A +a).

11. tan2 (45° — }; B) — tan 1⁄2 ( A − a) tan 1 (4+a).

§ 50. NAPIER'S RULES.

The ten formulas deduced in § 49 express the relations between five parts of a right triangle, the three sides and the two oblique angles. All these relations may be shown to follow from two very useful Rules, devised by Baron Napier, the inventor of logarithms.

For this purpose the right angle (not entering the formulas) is left out of account, and instead of the hypotenuse and the two oblique angles, their respective complements are employed; so that the five parts considered by the Rules are: a, b, co. A, co. c, co. B. Any one of these parts may be called a middle part; and then the two parts immediately adjacent are called adjacent parts, and the other two are called opposite parts.

Rule I. The sine of the middle part is equal to the product of the tangents of the adjacent parts.

Rule II. The sine of the 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.

Co. B

The correctness of these Rules may be shown by taking each of the five parts as middle part, and comparing the resulting equations with the equations contained in Formulas [38]-[43].

For example, let co. c be taken as middle part, then co. A and co. B are the adjacent parts, and a and b the opposite parts, as is very plainly seen in Fig. 41. Then, by Napier's Rules:

co.c

Co. A

FIG. 42.

α

results which agree with Formulas [38] and [43] respectively.

EXERCISE XXXI.

1. Show that Napier's Rules lead to the equations contained in Formulas [39], [40], [41], and [42].

2. What will 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 legs?

§ 51. SOLUTION OF RIGHT SPHERICAL TRIANGLES.

By means of Formulas [38]-[43] we can solve a right triangle in all possible cases. In every case two parts besides the right angle must be given.

CASE I. Given the two legs a and b.

The solution is contained in Formulas [38] and [42]; viz.:

cos c = cos a cos b,

tan Atan a csc b,

tan Btan b csc a.

=

For example, let a 27° 28′ 36", b=51° 12′ 8′′; then the solution by logarithms is as follows:

CASE II. Given the hypotenuse c and the leg a.

From Formulas [38], [39], and [40] we obtain

cos b = cos c sec a,

sin A sin a csc c,

cos Btan a cot c.