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Ex. 2. In a spherical triangle ABC there are given A=103° 59′ 57′′, B=46° 18′ 7′′, and a=42° 8′ 48′′; required the remaining parts.

There will but one triangle, since sin B<sin A.

CASE III.

For this case we use equations (3.).

Ans.

Having given the three sides of a spherical triangle to find the

angles.

cos A=R√sin § s sin (†s—a)

sin b sin c

Log sin
s 127° 11′ 30′′
log sin (s-a) 70° 31' 30"
-log sin b 83° 13'
-log sin c 114° 30'

Sum

Half sum =log cos A 24° 15', 39"

Ex. 1. In an oblique angled spherical triangle there are given a 56° 40', b=83° 13' and c-114° 30'; required the angles.

=

=127° 11' 30"

(a+b+c)=s
(b+c—a)=(s—a)=70° 31′ 30′′.

b = =30°
C=36° 7′ 54′′
c =24° 3′ 56′′.

ar.-comp.

ar.-comp.

Hence,

angle A=48° 31′ 18′′.

The addition of twice the logarithm of radius, or 20, to the numerator of the quantity under the radical just cancels the 20 which is to be subtracted on account of the arithmetical complements, to that the 20, in both cases, may be omitted.

Applying the same formulas to the angles B and C, we find,
B= 62° 55′ 46′′
C=125° 19′ 02′′.

Ans.

9.901250

9.974413

0.003051

0.040977

19.919691

9.959845

Ex. 2. In a spherical triangle there are given a—40° 18′ 29′′, b=67° 14′ 28′′, and c=89° 47′ 6′′ : required the three angles.

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CASE IV.

Having given the three angles of a spherical triangle, to find tre

three sides.

For this case we employ equations (7.)

cos a=RV

cos(S-B)cos(S—C)

sin B sin C

(S-A)
(S-B)

(S-C)

Ex. 1. In a spherical triangle ABC there are given A=48° 30′, B=125° 20′, and C=62° 54′; required the sides.

}(A+B+C)=}S= 118° 22′

69° 52'

6° 58'

55o 28'

Hence,
In a similar manner we find,

=

=

Log cos (S-B) —6° 58′
log cos (S-C) 55° 28'
-log sin B
-log sin

125° 20'

C

62° 54'

Sum

Half sum=log cos A=28° 19′ 48′′

ar.-comp.

ar.-comp.

side a 56° 39′ 36′′.

CASE V.

Ans.

9.996782

9.753495

0.088415

0.050506

19.889198

9.944599

Ex. 2. In a spherical triangle ABC, there are given A=109° 55′ 42′′, B=115° 38′ 33′′, and C=120° 43′ 37′′; required the three sides.

b=114° 29′ 58′′ c= 83° 12′ 06′′.

a= 98° 21′ 40′′ b=109° 50′ 22′′ c=115° 13′ 26′′

Having given in a spherical triangle, two sides and their included ingle, to find the remaining parts.

For this case we employ the two first of Napier's Analogies cos (a+b): cos (a-b): cot C: tang (A+B) sin (a+b) sin (a—b) :: cot C: tang (A-B).

Having found the half sum and the half difference of the angles A and B, the angles themselves become known; for, the greater angle is equal to the half sum plus the half difference, and the lesser is equal to the half sum minus the half diffe

rence.

The greater angle is then to be placed opposite the greater side. The remaining side of the triangle can then be found by Case II.

Ex. 1. In a spherical triangle ABC, there are given a=68° 46' 2", b=37° 10', and C-39° 23'; to find the remaining parts }(a+b)=52° 58′ 1′′, (a—b) = 15° 48′ 1′′, }C=19° 41′ 30′′. As cos (a+b) 52° 58′ 1" log.

ar.-comp.

Is to cos(a-b) 15° 48′ 1′′
So is cot C 19° 41' 30"
To tang (A+B) 77° 22′ 25′′

(a+b) 52° 58' 1" log. ar.-comp.
(a-b) 15° 48′ 1′′

As sin
Is to sin
So is cot C 19° 41′ 30′′

Totang (A-B) 43° 37′ 21′′

Hence, A=77° 22′ 25′′ +43° 37′ 21′′-120° 59′ 46′′
B=77° 22′ 25′′-43° 37′ 21′′ 33° 45′ 04′′
side c
= 43° 37' 37".

Ans.

0.220210

9.983271

10.446254

10.649735

=

CASE VI.

0.097840

9.435016

10.446254

9.979110

Ex. 2. In a spherical triangle ABC, there are given b=83 19′ 42′′, c=23° 27′ 46′′, the contained angle A=20° 39′ 48′ ; to find the remaining parts.

B=156° 30′ 16′′
C= 9° 11' 48"

a = 61° 32′ 12′′.

In a spherical triangle, having given two angles and the included side to find the remaining parts.

For this case we employ the second of Napier's Analogies. cos (A+B): cos (A—B) :: tang c: tang (a+b) sin (A+B) : sin † (A—B) : : tang c : tang (a—b). From which a and b are found as in the last case. The remaining angle can then be found by Case I.

}

Ex. 1. In a spherical triangle ABC, there are given A==81* 38′ 20′′, B=70° 9′ 38", c=59° 16' 23"; to find the remaining parts.

}(A+B)=75° 53′ 59′′, (A—B)=5° 44′ 21′′, c=29° 38′ 11′′. (A+B) 75° 53' 59" log. ar.-comp. 0.613287 (A-B) 5° 44′ 21′′

9.997818

9.755051

Ic 29° 38′ 11′′ (a+b) 66° 42′ 52′′

10.366158

As cos

To cos
So is tang

To tang

As sin

To sin

So is tang

To tang

Hence

(A+B) 75° 53' 59" log. ar.-comp. 0.013286
(A-B) 5° 14′ 21′′
c 29° 38' 11"

9.000000
9.755051

(a-b) 3° 21′ 25′′

8.768337

a=66° 42′ 52"+3° 21′ 25′′

70° 04′ 17′′

b=66° 42′ 52"-3° 21′ 25′′-63° 21′ 27′′
=64° 46′ 33".

angle C

Ex. 2. In a spherical triangle ABC, there are given A=34° 15′ 3′′, B=42° 15′ 13′′, and c=76° 35′ 36"; to find the remain

ing parts.

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274

MENSURATION OF SURFACES.

The area, or content of a surface, is determined by finding how many times it contains some other surface which is assumed as the unit of measure. Thus, when we say that a square yard contains 9 square feet, we should understand that one square foot is taken for the unit of measure, and that this unit is contained 9 times in the square yard.

The most convenient unit of measure for a surface, is a square whose side is the linear unit in which the linear dimensions of the figure are estimated. Thus, if the linear dimensions are feet, it will be most convenient to express the area in square feet; if the linear dimensions are yards, it will be most convenient to express the area in square yards, &c.

We have already seen (Book IV. Prop. IV. Sch.), that the term, rectangle or product of two lines, designates the rectangle constructed on the lines as sides; and that the numerical value of this product expresses the number of times which the rectangle contains its unit of measure.

PROBLEM I.

To find the area of a square, a rectangle, or a parallelogram. RULE.-Multiply the base by the altitude, and the product will be the area (Book IV. Prop. V.).

1. To find the area of a parallelogram, the base being 12.25 and the altitude 8.5. Ans. 104.125.

2. What is the area of a square whose side is 204.3 feet? Ans. 41738.49 sq. ft.

3. What is the content, in square yards, of a rectangle whose base is 66.3 feet, and altitude 33.3 feet? Ans. 245.31.

4. To find the area of a rectangular board, whose length is 12 feet, and breadth 9 inches. Ans. 93 sq. ft.

5. To find the number of square yards of painting in a par allelogram, whose base is 37 feet, and altitude 5 feet 3 inches. Ans. 21

PROBLEM II.

To find the area of a triangle.

CASE I.

When the base and altitude are given.

RULE.-Multiply the base by the altitude, and take half the product. Or, multiply one of these dimensions by half the other (Book IV. Prop. VI.).

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