Let the side BC of the quadrantal triangle BAC, be equal to 90°; produce the side CA till CD is equal to 90°, and conceive the arc of a great circle to be drawn through B and D.

Then C will be the pole of the arc BD, and the angle C will be measured by BD (B. IX.,

B

a

P. 4), and the angles CBD and D will be right angles. Now before the remaining parts of the quadrantal triangle can be found, at least two parts must be given in addition to the side BC = 90°; in which case two parts of the right-angled triangle BDA, together with the right angle, become known. Hence, the conditions which enable us to determine one of these triangles, will enable us also to determine the other.

EXAMPLES.

1. In the quadrantal triangle BCA, there are given CB=90°, the angle Ɑ = 42° 12', and the angle A = 115° 20′; required the remaining parts.

Having produced CA to D, making CD = 90°, and drawn the arc BD, there will then be given in the right angled triangle BAD, the side a C 42° 12', and the

=

-

angle BAD 180° - BAC 180° — 115° 20′ = 64° 40′, to find the remaining parts.

To find the side d.

The side a is the middle part, and the extremes opposite: hence,

The angle A corresponds to the middle part, and the extremes are opposite: hence,

The side is the middle part, and the extremes are

CBA = 90° - ABD = 90° - 35° 16′ 53′′ = 54° 43′ 07'

BA = d

2. In the right-angled triangle BAC, right-angled at A, there are given a = 115° 25', and c = 60° 59′: required

3. In the right-angled spherical triangle BAC, rightangled at A, there are given c = 116° 30′ 43′′, and b = 29° 41′ 32" required the remaining parts.

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4. In a quadrantal triangle, there are given the quadrantal side 90°, an adjacent side 115° 09', and the included angle 115° 55': required the remaining parts.

=

SOLUTION OF OBLIQUE-ANGLED TRIANGLES BY LOGARITHMS,

18. There are six cases which occur in the solution of oblique-angled spherical triangles.

1. Having given two sides, and an angle opposite ɔne of them.

2. Having given two angles, and a side opposite one of them.

3. Having given the three sides of a triangle, to find the angles.

4. Having given the three angles of a triangle, to find the sides.

5. Having given two sides and the included angle.
6. Having given two angles and the included side.

CASE I.

Given two sides, and an angle opposite one of them, to find the remaining parts.

19. For this case, we employ proportions (3);

sin a : sin b :: sin A sin B.

Ex. 1. Given the side a = 44° 13′ 45′′, b 84° 14' 29", and the angle A= 32° 26' 07": required the remaining parts.

To find the angle B.

b

A

B

C

B'

sin α 44° 13' 45" : sin b 84° 14' 29" :: sin A 32° 26' 07'

ar. comp. log.

0.156437

9.997803

9.729445

: sin

B 49° 54′ 38′′, or sin B' 130° 5′ 22′′ 9.883685

Since the sine of an arc is the same as the sine of its supplement, there are two angles corresponding to the logarithmic sine 9.883685, and these angles are supple ments of each other. It does not follow, however, that both of them will satisfy all the other conditions of the question. If they do, there will be two triangles ACB', ACB; if not, there will be but one.

To determine the circumstances under which this ambiguity arises, we will consider the 2d of equations (1) cos b = cos a cos c + sin a sin c cos B,

from which we obtain,

cos B =

cos b COS a COS C

sin a sin c

Now, if cos b be greater than cos a, we shall have,

cos b> cos a cos'c,

or, the sign of the second member of the equation will depend on that of cos b. Hence, cos B and cos b will have the same sign, or B and b will be of the same species, and there will be but one triangle.

But when cos b> cos a, then sin b < sin a: hence, If the sine of the side opposite the required angle be less than the sine of the other given side, there will be but one triangle. If, however, sin b> sin a, the cos b will be less than cos a, and it is plain that such a value may then be given to c, as to render

cos b < cos a cos c,

or, the sign of the second member may be made to depend

on cos c.

We can therefore give such values to c as to satisfy the two equations,

hence, if the sine of the side opposite the required angle be greater than the sine of the other given side, there will be two triangles which will fulfil the given conditions.

Let us, however, consider the triangle ACB, in which we are yet to find the base AB and the angle C. We can find these parts by dividing the triangle into two rightangled triangles. Draw the arc CD perpendicular to the base AB: then, in each of the triangles there will be given. the hypothenuse and the angle at the base. And generally,

when it is proposed to solve an oblique-angled triangle by means of the right-angled triangle, we must so draw the perpendicular, that it shall pass through the extremity of a given side, and lie opposite to a given angle.

:

To find the angle C, in the triangle ACD.

9.803105

10.000000.

9.001465

8.804570

To find the angle C in the triangle DCB.

The arc 64° 43' 48", which corresponds to sin c is not the value of the side AB: for the side AB must be greater than b, since it lies opposite to a greater angle. But b 84° 14′ 29′′: hence, the side AB must be the supplement of 64° 43′ 48", or, 115° 16′ 12′′.

=

Ex. 2. Given b 91° 03′ 25", a 40° 36' 37", and A = 35° 57′ 15′′: required the remaining parts, when the obtuse angle B is taken.