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Extend on the line of sines from 90° to 65° 32', the supplement of 114° 28', and that extent will reach in the same direction from 88° 4' to about 6510 on the line of sines. Extend from 652° to 18° 11', the supplement of 161° 49', and that extent will reach, in the same direction, from 40° 43' to about 13° on the line of sines, below which stands on the line of versed sines nearly 1234, the required angle.

APPLICATION OF TRIGONOMETRY TO THE NUMERICAL COMPUTATION OF THE DIFFERENT PARTS OF RIGHT ANGLED PLANE TRIANGLES.

That all the parts of a right angled plane triangle may become determinate, three parts of it must be given; and, at least, one of these parts must be a side. Now the right angle is always a given quantity, therefore either two of the sides, or a side and one of the acute angles, must be given, before the other parts can be computed.

In Prop. 3. of the Theory of Trigonometry, the following proportions have been deduced, which are sufficient for the solution of any case that can arise from considering different parts of the triangle as given.

1. Any side of a triangle is to the sine of its opposite angle, as any other side is to the sine of its opposite angle; and conversely, the sine of any angle is to its opposite side, as the sine of any other angle is to its opposite side.

2. Either of the sides containing the right angle is to radius as the other side is to tangent of its opposite angle; and, conversely, radius is to either of the sides about the right angle, as the tangent of the angle opposite the other side is to that other side.

3. Either of the sides containing the right angle is to radius as the hypothenuse is to the secant of the acute angle adjacent to that side; and, conversely, radius is to either of the sides containing the right angle as the secant of the acute angle adjacent to that side is to the hypothenuse.

Remark. As each of the acute angles is to the complement of the other, the cosine, cotangent, or cosecant of either of the angles may be substituted, in the above proportions, for the sine, tangent, or secant of the other.

The proportions will in general be most conveniently solved by logarithms; the logarithm of the first term taken from the sum of the logarithms of the second and third will leave the logarithm of the required result.

To the above proportions the following useful properties of right angled plane triangles may be added.

1. The number representing the hypothenuse is equal to the square root of the sum of the squares of the numbers representing the other two sides. (Geo. Theo. 44 )

2. The number representing either of the sides is equal to the square root of the product of the sum, and difference of the numbers representing the hypothenuse and the other side. (Geo. Theo. 44. Cor.)

EXAMPLE I. In the right angle triangle A B C, given the hypothenuse AC 840:4, and the 2 A 38° 16', to find the other parts. .

BY CONSTRUCTION. Draw the horizontal line A B, and with the aid of the line of chords, make the angle A equal to about 3810, on AC

с from a scale of equal parts lay off 840:4, and from Clet fall on A B the perpendicular B C, then A B and BC measured on the scale of equal parts from which AC

A

B was taken, will be found respectively to be about 660, and 520.

By CALCULATION,
The angle C being the complement of the angle A, is 51° 44'.

To find A B.
As radius

10.000000
: A C 840•4...

2:924486 : : coş A 38' 16'

9.894945

12 819431 : A B 659.8...

2.819431

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BY GUNTER'S SCALE. Extend on the line of sines, from rad or sine 90°, to 38° 16'; that extent applied on the line of numbers, from 840•4 towards the left, will reach to B C, 520.5. Again extend on the line of sines from rad to 51° 44', the complement of the angle A, and that extent applied on the line of numbers from 840:4 towards the left,

or

BY INSPECTION. In Table 2, under 38°, and opposite 210, the fourth part of AC, in the column of dist., we have 165.5, for the fourth part of A B, in the column of lat., and 129.3 for the fourth part of B C, in the column of dep.

And under 39°, and opposite 210, we have 163.2 and 132-2 for the fourth part of A B and B C respectively. Hence with the angle A 38° the fourth part of A B is 165:5, and of B C 129.3 39°

1632

132:2 Difference for one degree 2.3

+ 2.9. Now the difference between 38o and the given angle is nearly a A B

2.3 quarter of a degree. Hence is nearly equal to 165.5 – 4

4

В С – , 164

is

4 2:9 pearly equal to 129.3 + or to 129.3 + 7 or to 130 ; or BC = 520-0 nearly

EXAMPLE II. In the right angle triangle A BC, (see the last figure) given the base A B 1214, and the angle A 51° 40' 30" to find the other parts.

BY CONSTRUCTION. Draw the horizontal line A B, and make it of the proposed length from any convenient scale of equal parts. Make the angle A of the given magnitude, and draw B C perpendicular to A B, meeting A C in C. Then A C, measured on the scale of equal parts from which AC was taken, will be found 1958, and B C, measured on the same scale, will be found 1536 nearly.

By CALCULATION.

To find A C.
As radius

10.000000
: A B 1214 ....

3.084219 : : sec 4 A 51° 40' 30" 10:207523

13:291742 : A C 1957

3.291742

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Extend on the line of sines from 38° 19' 30'', the complement of the angle A, to radius, or sine of 90°; that extent applied on the line of numbers from 1214 towards the right will reach to A C 1958. Again extend from radius, or 45°, on the line of tangents, to 51° 40 30%, that extent applied on the line of numbers from 1214 towards the right will reach to B C 1536.

BY INSPECTION.

With the angle A considered as 51°, at the bottom of Table 2, and 121:4, the tenth part of A B, in the coluinn of lat. we have in the columns of dist, and dep. about 193, and 150 for the corresponding parts of A C and B C. And taking 52° for the angle A, we have 197 and 155.2 nearly, for the tenth parts of A C and B C. Hence with the angle A 51° we have A C 193 and B C 150

520

197

155.2 Difference for one degree + 4

+5.2

Now 40' 30" being about į of a degree, we have the tenth part of AC = 193 + f of 4 nearly, = 193 + 2:6 nearly = 195:6 nearly; whence A C = 1956 nearly. And the tenth part of B C = 150 + of 5•2 = 150 + 3:4 = 153:4; whence BC = 1534 nearly.

EXAMPLES FOR EXERCISE.

Let A B C (Fig. p. 110) represent any plane triangle right angled at B. 1. Given A C 73:26, 4 A 49° 12' 20", required the other parts ?

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2. Given A B 469:34, and 2 A 51° 26' 17", to find the other parts?

Answer, 2 C 38° 33' 43", B C 588:5, and A C 752:9. 3. Given A B 493, and 4 C 20° 14', required the remaining parts?

Answer, 4 A 69° 46', B C 1338, and A C 1425. 4. Let A B = 331, A = 49° 14', what are the other parts ?

Answer, AC 506:9, B C 383.9, and < C 40° 46'. 5. If A C = 45, and 2 C = 37° 22', what are the remaining parts?

Answer, A B 27:31, B C 3576, and 4 A 52° 38'. 6. Given A C 4264:3, and 4 A 56° 29' 13', to find the remaining parts ?

Answer, A B 2354:4, B C 3555:4, and 2 C 33° 30°47'. 7. If A B = 42:2, and 2 A = 31° 12' 49', what are the other parts?

Answer, AC 51.68, B C 26:78, and 2 C 58° 47' 11'. 8. If A B = 8372:1, and BC= 694:73, what are the other parts ?

Answer, A C 8400:9, 2 C 85° 15', and 2 A 4° 45'. 9. If A B be 63.4, and A C be 8572, what are the other parts ?

Answer, B C 5700g2 C 47° 42, and A 42° 18'. 10. Given A C 1269, and A B 3162, to find the other parts ?

Answer, B C 6546, 2 C 25° 47'7", and L A 64° 12' 53". 11. Given A C 4824, and B C 2412, to find the other parts ?

Answer, L A 30° 00', 2 B 60° 00', and A B 4178. 12. If A B = 621:3, and AC = 918:4, what are the remaining parts?

Answer, LA 47° 26', / C 42° 34', and B C 676.5. 13. If AC = 4184, and A B = 2632, what are the remaining parts ?

Answer, B C 3252:6, _ A 51° 1' 13", and 4 C 38° 58' 47". 14. If A B = 732:1, and ZA = 60° 10, what are the other parts?

Answer, AC 1471:6, B C 1276.59, and 4 C 29° 50'. 15. Given A C and B C, the first 90 and the second 70, to find the other parts?

Answer, A B 56 58, L A 51° 3', and 2 C 38° 57'. 16. Given A B = 416:3, and A C = 820:4, to find the remaining parts ?

Answer, B C 7064, A 59° 30', and Z C 300 30'. 17. Let A C be 225, and B C be 90:37, required the other parts ?

Answer, A B 206, 2 A 23° 41', and 2 C 66° 19. 18. Given Z A 70° 4' 38, and A B 67289, to find the remaining parts ?

Answer, B C 185653, AC 197471, and 2 C 19° 55' 22". 19. Given < C 88° 4' 56', and A C and 7396-2, to find the remaining parts ?

Answer, B C 247-2, A B 7391.93, and 4 A 1° 55' 4". 20. Given A C 631.25, B C 74:9, to find the other parts ?

Answer, A B 626:8, 4 A 6° 48' 52', and 4 C 83° 11' 8".

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