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c

d

right angled at A', and hence every case may be referred to a right angled triangle.

But we can solve the quadrantal triangle by means of the right angled triangle in a manner still more simple.

In the quadrantal triangle BAC, in which BC=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

A the angle will be measured by B BD (Book IX. Prop. VI.), and the

16 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.

3. In the quadrantal triangle BCA, there are given CB=90°, the angle C=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 BĂD=180°— BAC=180°—115° 20'=64° 40', to find the remaining parts.

To find the side d. The side a will be the middle part, and the extremes oppo.. site: hence,

R sin a=sin A sin d. As sin A 64° 40'

ar.-comp. log. 0.043911 Is to R

10.000000 So is sin a 42° 12'

9.827189 To sin d 48° 00' 14"

9.871100

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To find the angle B. The angle A will correspond to the middle part, and the extremes will be opposite : hence

Rcos A=sin B cos a. As cos

42° 12' ar.-comp. log. 0.130296 R

a

10.000000 So is cos A

64° 40'
°

9.631326 To sin B 35° 16' 53"

9.761622

Is to

To find the side b. The side b will be the middle part, and the extremes adjacent: hence,

R sin b=cot A tang a. As R

ar.-comp.

log.

0.000000 Is to cotA 64° 40'

9.675237 So is tang a 42° 12'

9.957485 To sin b 25° 25' 14"

9.632722

Hence, CA=90o--b=90°—25° 25' 14" =64° 34' 46"

CBA=90°_ABD=90°-35° 16' 53"=54° 43'07"
BA=d

=48° 00' 15". 4. In the right angled triangle BAC, right angled at A, there are given a=115° 25', and c=60° 59' : required the remaining parts.

B=148° 56' 45" Ans. C= 75° 30' 33"

b=152° 13' 50". 5. In the right angled spherical triangle BAC, right angled at A, there are given c=116° 30' 43", and b=29° 41' 32" : required the remaining parts.

C-103° 52' 46" Ans. B- 32° 30' 22''

a =112° 48' 58". 6. 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.

side, 113° 18' 19" Ans.

117° 33' 52" 101° 40' 07.

angles, {

SOLUTION OF OBLIQUE ANGLED TRIANGLES BY LOGARITHMS.

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

1. Having given two sides, and an angle opposite one 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.

z

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

5. Ilaving 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 re

maining parts.

С

For this case we employ equation (1.);

As 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

6 the remaining parts.

A.
To find the angle B.
As sin 44° 13' 45" ar.-comp.

log
Is to sin b 84° 14' 29"
So is sin A 32° 26' 07"
To sin B 49° 54' 38" or sin B' 130° 5' 22"

B

B' D

a

0.156437 9.997803 9.729445 9.883685

Since the sine of an arc is the same as the sine of its supple. ment, there will be two angles corresponding to the logarithmic sine 9.883685 and these angles will be supplements 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 (2.).

R2 cos b=R cos a cos c+sin a sin c cos B. from which we obtain

R? cos Rcos a cos c

sin a sin c Vow if cos b be greater than cos a, we shall have

R? cos b>R 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 6 will have the same

cos B

sign, or B and b will be of the same species, and there will be but one triangle.

But when cos b>cos a, 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

R2 cos b<R cos a cos C, or the sign of the second member may be made to depend on

COS C.

cos B

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

R? cos b-R cos a cos c
+cos B=

sin a sin c
R? cos b-R cos a cos c

sin a sin c 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 most readily by dividing the triangle into two right angled triangles. Draw the arc CD perpendicular to the base AD : 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 10site to a given angle.

To find the angle C, in the triangle ACD.
As cot
A 32° 26'07"

ar.-comp. log. 9.803105 R

10.000000 So is cos b 84° 14' 29"

9.001405 To cot ACD 86° 21' 09"

8.804570

ls to

To find the angle C in the triangie DCB. As cot B 49° 54' 38" ar.-comp. log.

0.074310 Is to R

10.000000 So is cos 44° 13' 45''

9.855250 To cot DCB 49° 35' 38"

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9.930060

Hence

ACB=135° 56' 47".

log.

To find the side AB.
As sin A 32° 26'07" ar.-comp.
Is to sin C 135° 56' 47"
So is sin 44° 13' 45'
To sin c 115° 16' 29"

0.270555 9.842191 9.843563 9.956309

a

The arc 64° 43' 31", 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' 31", or 115° 16' 29".

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.

B=115° 35' 41" Ans. C= 58° 30' 57"

70° 58' 52"

C

CASE II.

Having given two angles and a side opposite one of them, to find

the remaining parts.

For this case, we employ the equation (1.)

sin A : sin B :: sin a : sin b. Ex. 1. In a spherical triangle ABC, there are given the angle A=50° 12', B=58° 8', and the side a=

=62° 42'; to find the remaining parts.

To find the side b.

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As sin A 50° 12' ar.-comp.

log.
Is to sin B 58° 08'
So is sin a 62° 42'
To sin b 79° 12' 10", or 100° 47' 50"

0.114478 9.929050 9.948715 9.992243

We see here, as in the last example, that there are two arcs corresponding to the 4th term of the proportion, and these arcs are supplements of each other, since they have the same sine. It does not follow, however, that both of them will satisfy all the conditions of the question. If they do, there will be two triangles; if not, there will be but one.

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