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SOLUTION OF OBLIQUE-ANGLED TRIANGLES BY LOGARITEMS.

18. There are six cases which occur in the solutioof 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.

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.

sin a

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a

Given two sides, and an angle opposite one of them, to find

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

sin b sin A : sin B.
Ex. 1. Given the side a =
44° 13' 45", b = 84° 14' 29",

6 and the angle A = 32° 26'07" : required the remaining parts.

A

B

D To find the angle B. sin 44° 13' 45" ar. comp. log. 0.156437 sin b. 84° 14' 29"

9.997803 :: sin A 32° 26' 07"

9.729445 sin B 49° 54' 38", or sin B' 130° 5' 22" 9.883685

a

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B

(L

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:

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,

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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 6 > 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 6 > 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

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

COS

on COS C.

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

cos b - cos a cos C + cos B=

sin a sin c

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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 Vase 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. cot A 32° 26' 07" ar. comp. log.

9.803105 : R

10.000000 :: COS 5 84° 14' 29".

9.001465 : cot ACD 86° 21' 06"

8.804570

.

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

0.074810 R

10.000000 44° 13' 45".

9.855250 : cot DCB 49° 35' 38".

9.930060

.

COS

a

Hence,

ACB = 135° 56' 44".

To find the side AB.

sin A 32° 26'07" ar. comp. log. i sin C 135° 56' 44".

44° 13' 45'. : sin c 115° 16' 12" .

0.270555 9.842198 9.843563 9.956316

:: sin

a

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.

B = 115° 35' 41" Ans. C= 58° 30' 57"

70° 58' 52"

CASE II.

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

find the remaining parts.
20. For this case, we employ the proportions (3).

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';

A' to find the remaining parts.

B A
To find the side b.
sin A 50° 12 ar, comp.

log. 0.114478
sin
B 58° 08' .

9.929050 62° 42'

9.948715 5 79° 12' 10", or, 100° 47' 50" 9.992243 We see here, as in the last example, that there are two angles corresponding to the 4th term of the proportion, and these angles 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

:

a

:: sin : sin

but one.

To determine when there are two triangles, and also when there is but one, let us consider the second of equations (2), cos B = sin A sin C cos 6 – cos A cos C,

cos B + cos A cos C which gives, cos b =

sin A sin c Now, if cos B be greater than cos A, we shall have,

cos B > cos A cos C, and hence, the sign of the second member of the equation will depend on that of cos B, and consequently cos b and cos B will have the same algebraic sign, or b and B will be of the same species. But when cos B > cos A the sin B < sin A: hence,

If the sine of the angle opposite the required side be less than the sine of the other given angle, there will be but one solution.

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 cos C, as to render

cos B < cos A cos C, or, the sign of the second member of the equation may be made to depend on cos C. We can therefore give such values to C as to satisfy the two equations,

cos B + cos A cos C + cos b =

sin A sin c

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and

cos 6

cos B + cos A cos C

sin A sin C

Hence, if the sine of the angle opposite the required side be greater than the sine of the other given angle, there will be two solutions.

Let us first suppose the side b to be less than 90°, or, equal to 79° 12' 10".

If, now, we let fall from the angle C, a perpendicular on the base BA, the triangle wil be divided into two rightangled triangles, in each of which there will be two parts known besides the right angle. Calculating the parts by Napier's rules, we find,

C= 130° 54' 28"

c= 119° 03' 26" If we take the side b = 100° 47' 50", we shall find,

C= 156° 15' 06"

c= 152° 14' 18" Ex. 2. In a spherical triangle ABC, there are given A = 103° 59' 57", B = 46° 18' 07", and a = 42° 08' 48"; required the remaining parts. There will be but one triangle, since sin B < sin A.

b= 30° Ans. C= 36° 07' 54"

c= 24° 03' 56'

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