Page images
PDF
EPUB

therefore, the side DB, which is the complement of this, is 64° 34' 40".

2. In the triangle DAB, DA=90°, A=112° 2′ 9′′, and AB= 67° 3′ 14′′, to find the other parts.

11

Since in this example A is obtuse, DB is obtuse. In the right-angled triangle ABC we have A= 22° 2′9′′ and AB=67° 3′ 14′′; let A be the middle part, then AB, AC, will be adjacent parts, and we shall have

rad. X sin. comp. A tan. b x tan. comp. c;

that is,

rad. cos. A tan. b cot. c .. tan. b

D

A

rad. cos. A

[blocks in formation]

therefore, the angle D65° 27′ 9′′.

Take now a for the middle part, then A and c will be opposite parts;

[blocks in formation]

and a will be acute, because the opposite angle is acute

[blocks in formation]

As we have now to find B, take a for the middle part, then b and B will be adjacent parts, therefore

[blocks in formation]

3. Given the quadrantal side and the other two sides equal to 22° 53′ 30′′, and 51° 4′ 35′′, to find the angle opposite to the quadrantal side.

B=70° 3′ 44".

4. In the quadrantal triangle ADB are given D=69° 13′ 46′′, and A=72° 12′ 4′′, to determine the other parts.

AB=70° 8′ 39′′, BD = 73° 17′ 29′′, B=96° 13′ 23′′.

These examples will suffice, for the present, to show the application of Napier's rules to the solution of right-angled and quadrantal triangles. We shall, therefore, now give examples of the solution of the various cases of oblique angled triangles in general.

Solution of Oblique Angled Spherical Triangles.

(55.) The fundamental equations (A) show that in order to determine the several parts of a spherical triangle, three of those parts must be previously given. Now, three parts out of the six can be combined only in these different ways, viz.

1. The three sides.

2. The three angles.

3. Two sides and the included angle.

4. Two angles and the interjacent side.

5. Two sides and an opposite angle.

6. Two angles and an opposite side.

So that the complete solution of an oblique angled spherical triangle presents six cases. These we shall solve in the order in which they are here enumerated.

CASE I.

(56.) Given the three sides to find the angles.

For the determination of any angle A we have by (47) the three following different expressions, viz.

[blocks in formation]

tan. § A = √ sin. (3 S − b ) sin. ( § S — c)
sin. S sin. ( S − a)

We may apply to these formulas the remarks made at (21) in the Plane Trigonometry. It will be sufficient to observe here that the first formula is generally the most suitable, because the angle A is rarely so large as to be very near 180°.

EXAMPLES.

1. In an oblique spherical triangle the three sides are

a=68° 46′ 2′′, b = 43° 37′ 38′′, c = 37° 10′;

required the angle A.

[blocks in formation]

2. Given a 108°, b = 52° 12′, and c = 74° 30', to find A.

[blocks in formation]

3. Given a = 70° 4′ 18′′, b=63° 21′ 27′′, and c = 59° 16′ 23′, to

find the angles A and B.

A=81° 38′ 20′′, B=70° 9′ 38′′.

4. Given a=67° 25′ 2′′,b = 80° 2′ 25′′, c=23° 27′ 46′′, to find the angle A.

A=54° 55′ 19′′.

=61° 32′ 12′′, b⇒ 83° 19′ 42′, c = 23° 27′ 46′′, to find A.

5. Given a =

CASE II.

A 20° 39′48′′.

(57.) Given the three angles to find the sides.

By (49) we have the following formulas for any side a in terms of the three angles, viz.

[blocks in formation]

It may be remarked here that the first two only of the expressions in this and in the former case need be borne in the memory, as the third is an immediate consequence of them. If the expressions in the former case be recollected, these can scarcely fail to be recalled at the same time, as they differ from them only in this, viz. that the sides are replaced by their opposite angles, and, except in the denominators, cosines are written for sines, and sines for cosines.

EXAMPLES.

1. The three angles of a spherical triangle are,

A = 130° 3′ 11′′, B=31° 34′ 26′′, C=30° 28′ 12′′,

required the side a.

« PreviousContinue »