1. Geometrically. Draw an indefinite line, upon which set off AB = 345, from some convenient scale of equal parts.—Make the angle A = 37°4.—With a radius of 232, taken from the same scale of equal parts, and centre B, cross AC in the two points C, C.-Lastly, join BC, BC, and the figure is constructed, which gives two triangles, showing that the case is ambiguous. Then, the sides AC measured by the scale of equal parts, and the angles B and C measured by the line of chords, or other instrument, will be found to be nearly as below; viz. AC 174 or 3744 angle B 27° or EXAMPLE II. In the plane triangle ABC, Given, angle A 57° 12′ 24 45 angle C 115° or 3. Instrumentally. In the first proportion.-Extend the compasses from 232 to 345 upon the line of numbers; then that extent will reach, on the sines, from 37° to 64°, the angle C. In the second proportion.-Extend the compasses from 37° to 27° or 78°4, on the sines; then that extent will reach, on the line of numbers, from 232 to 174 or 374, the two values of the side AC. log. 2.3654880 9-7827958 2-5378191 9.9551269 Ans. log. 9-7827958 2.3654880 9.6580371 9.9908291 2.2407293 2.5735213 Then it will be, Ans. As the sum of those two sides, Is to the difference of the same sides; THEOREM IL When two sides and their contained angle are given. (angle B 64° 34′ 21′′ or, 7 7 21 112.65 feet 16:47 feet Demonstr.-Let ABC be the proposed triangle, having the two given sides AC, BC, including the given angle C. With the centre C, and radius CA, the less of these two sides, describe a semicircle, meeting the other side BC produced in D and E. Join AE, AD, and draw DF parallel to AE. or, AB E So is the tang. of half the sum of their opposite angles, To the tang. of half the difference of the same angles. Hence, because it has been shown under Algebra, that the half sum of any two quantities increased by their half difference, gives the greater, and diminished by it gives the less, if the half difference of the angles, so found, be added to their half sum, it will give the greater angle, and subtracting it will leave the less angle. Then, all the angles being now known, the unknown side will be found by the former theorem. D BD the difference of Then, BE is the sum, and the two given sides CB, CA. Also, the sum of the two angles CAB, CBA, is equal to the sum of the two CAD, CDA, these sums being each the supplement of the vertical-angle C to two right angles: but the two latter CAD, CDA, are equal to each other, being opposite to the two equal sides CA, CD: hence, either of them, as CDA, is equal to half the sum of the two unknown angles CAB, CBA. Again, the exterior angle CDA is equal to the two interior angles B and DAB; therefore the angle DAB is equal to the difference between CDA and B, or between CAD and B; consequently the same angle DAB is equal to half the difference of the unknown angles B and CAB; of which it has been shown that CDA is the half sum. Now the angle DAE, in a semicircle, is a right angle, or AE is perpendicular to AD; and DF, parallel to AE, is also perpendicular to AD: consequently, AE is the tangent of CDA the half sum, and DF the tangent of DAB the half difference of the angles, to the same radius AD, by the definition of a tangent. But, the tangents AE, DF, being parallel, it will be as BE: BD :: AE: DF; that is, as the sum of the sides is to the difference of the sides, so is the tangent of half the sum of the opposite angles, to the tangent of half their difference. Note. The sum of the unknown angles is found, by taking the given angle from 180°. 1. Geometrically. = Draw AB Set off AC Make the angle A = 37° 20. Join BC, and it is done. 345 from a scale of equal parts. 174 by the scale of equal parts. Then the other parts being measured, they are found to be nearly as follows; viz. the side BC 232 yards, the angle B 27°, and the angle C 115o). 2. Arithmetically. As sum of sides AB, AC, ......... So tangent half sum angles C and B, EXAMPLE I. Given, Their sum gives angle C Then, by the former theorem, As sine angle C 115° 36′ or 64° 24/ In the plane triangle ABC, angle A Required the other parts. AB 365 poles AC 154-33 In the plane triangle ABC, Given, AC 120 yards BC 112 yards angle C 57° 58' 39' Required the other parts. EXAMPLE II. EXAMPLE III 519-07 log. 2.7152259 2.2328183 10-4712979. 9-9888903 115 36 3. Instrumentally. In the first proportion.-Extend the compasses from 519 to 171, on the line of numbers; and that extent will reach, on the tangents, from 71° (the contrary way, because the tangents are set back again from 45°), a little beyond 45, which being set so far back from 45, falls upon 44°4, the fourth term. In the second proportion. — Extend from 64° to 37°, on the sines; and that extent will reach, on the numbers, from 345 to 232, the fourth term sought. Ans. ... Ans. 3 ... ... log. 9.9551259 2.5378191 2.3654890 BC 309-86 angle B 24° 45' angle C 98° 3 AB 112.65 angle A 57° 27′ 0 angle B 64 34 21 THEOREM III. When the three sides of the triangle are given. Then, having let fall a perpendicular from the greatest angle upon the орро site side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles; it will be, As the base, or sum of the segments, Is to the sum of the other two sides; So is the difference of those sides, To the difference of the segments of the base. Then half the difference of the segments being added to the half sum, or the half base, gives the greater segment; and the same subtracted gives the less segment. Hence, in each of the two right-angled triangles, there will be known two sides, and the angle opposite to one of them; consequently, the other angles will be found by the first problem. Demonstr.-By Cor. to Theorem 35, Geometry, the rectangle under the sum and difference of the two sides, is equal to the rectangle under the sum and difference of the two segments. Therefore, by forming the sides of these rectangles into a proportion, it will appear that the sums and differences are proportional as in this theorem, by Theor. 76, Geometry. To find the angles. 1. Geometrically. Draw the base AB centre A, describe an arc; arc, cutting the former in C. Then, by measuring the angles, they will be found to be nearly as follow; viz. angle A 27o, angle B 37°, and angle C 115o§. 345 by a scale of equal parts. With radius 232, and and with radius 174, and centre B, describe another Join AC, BC, and it is done. 2. Arithmetically. Having let fall the perpendicular CP, it will be, As the base AB: AC + BC :: AC- BC: AP BP, its half is 34.09 90° 206.59 62° 56'.. 90 00 27 04 ..... 206·59 = AP BP log. 2.3654880 10-0000000 2.3151093 9.9496213 Again, in the triangle BPC, right-angled at P, As the side BC To sine opposite angle P 174 07 To sin. opposite angle BCP... 52° 40′ 90 00 Leaves the angle B ........ 37 20 Also, the 62° 56' angle ACP Added to angle BCP 52 40 Gives the whole angle ACB 115 36 So that all the three angles are as follow; viz. the angle A 27° 4′; the angle B 37° 20′; the angle C 115° 36'. 3. Instrumentally. ...... ...... ...... In the plane triangle ABC, AB 365 poles Given the sides, AC 154.33 BC 309-8€ To find the angles. In the plane triangle ABC, AB 120 Given the sides, AC 112-65 BC 112 To find the angles. In the first proportion.-Extend the compasses from 345 to 406, on the line of numbers; then that extent will reach, on the same line, from 58 to 68.2 nearly, which is the difference of the segments of the base. In the second proportion.-Extend from 232 to 206, on the line of numbers; then that extent will reach, on the sines, from 90° to 63°. In the third proportion. — Extend from 174 to 138; then that extent will reach from 90° to 52° on the sines. EXAMPLE II. Ans. Ans. log. 2.2407239 10-0000000 9-9004436 As radius, i. e. sine of 90° or tangent of 45° The three foregoing theorems include all the cases of plane triangles, both right-angled and oblique; besides which, there are other theorems suited to some particular forms of triangles, which are sometimes more expeditious in their use than the general ones; one of which, as the case for which it serves so frequently occurs, may be here taken, as follows: So is the tangent of its adjacent angle To the other leg; angle A 57° 12 THEOREM IV. When, in a right-angled triangle, there are given one leg and the angles; to find the other leg or the hypothenuse; it will be, And so is the secant of the same angle angle A 57° 27′ 00′′ angle B 57 58 39 angle C 64 34 21 1 |