AC, BC, or of CD, CB. Also, the external angle ACE, is equal to the given sum of the two internal angles CAB, CBA; but the angle ADE, at the circumference, is equal to half the angle ACE at the centre; therefore the same angle ADE is equal to half the given sum of the angles CAB, CBA. Also, the external angle AGC, of the triangle BCG, is equal to the sum of the two internal angles GCB, GBC, or the angle GCB is equal to the difference of the two angles AGC, GBC; but the angle CAB is equal to the said angle AGC, these being opposite to the equal sides AC, CG; and the angle DAB, at the circumference, is equal to half the angle DCG at the centre; therefore the angle DAB is equal to half the difference of the two angles CAB, CBA; of which it has been shown that ADE or 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 tan gent. 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. Draw AB = 345 from a scale of equal parts. Make the angle A = 37° 20'. Set off AC = 174 by the scale of equal parts. Join Bc, and it is done. Then the other parts being measured, they are found to be nearly as follow; viz. the side BC 232 yards, the angle B 27°, and the angle c 115°4. So tang. half sum Zs cand в 71° 20 2.232818 10-471298 9-988890 To tang. half diff. scand B 44 16 - these added give ∠ c 115 36 Then, by the former theorem, In the first proportion.-Extend the compasses from 519 to 171, on the line of numbers; then that extent reaches, 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; then that extent reaches, on the numbers, from 345 to 232, the fourth term sought. When the Three Sides of a Triangle are given. First, let fall a perpendicular from the greatest angle on the opposite side, or base, dividing it into two segments, and the whole triangle into two right-angled triangles: then the proportion will be, As As the base, or sum of the segments, To the diff. of the segments of the base. Then take half this difference of the segments, and add it to the half sum, or the half base, for the greater segment; and subtract the same for the less segment. Hence, in each of the two right-angled triangles, there will be known two sides, and the right angle opposite to one of them; consequently the other angles will be found by the first theorem. Demonstr. By theor. 35, Geom. the rectangle of the sum and difference of the two sides, is equal to the rectangle of the sum and difference of the two segments. Therefore, by forming the sides of these rectangles into a proportion by theor. 76, Geometry, it will appear that the sums and differences are proportional as in this theorem. Draw the base AB = 345 by a scale of equal parts. With radius 232, and centre A, describe an arc; and with radius 174, and centre B, describe another arc, cutting the former in c. Join AC, BC, and it is done. Then, by measuring the angles, they will be found to be nearly as follows, viz. ZA 27°, ∠B 37°, and c115. 2. Arithmetically. Having let fall the perpendicular cp, it will be, BP, that is, as 345:406-07::57.93: 68.18 = AP - BP, BP and their diff. is 13841 Then, Then, in the triangle APC, right-angled at F, Again, in the triangle BPC, right angled at P, gives the whole / ACB 115 36 So that all the three angles are as follow, viz. the ∠ A 27° 4'; the ∠ в 37° 20'; the 2 c115° 36′. 3. Instrumentally. In the first proportion.-Extend the compasses from 345 to 406, on the line of numbers; then that extent reaches, 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 reaches, on the sines, from 90° to 63°. In the third proportion.-Extend from 174 to 1384; then that extent reaches from 90° to 52° on the sines. The three foregoing theorems include all the cases of plane triangles, both right-angled and oblique. But 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: THEOREM IV. When a Triangle is Right-angled; any of the unknown parts may be found by the following proportions: viz. Demonstr. AB being the given leg, in the right-angled triangle ABC; with the centre A, and any assumed radius AD, describe an arc De, and draw DF perpendicular to AB, or parallel to вс. Then it is evident, from the definitions, that DF is the tangent, and AF the secant of the arc DE, or of the angle a which is measured by that arc, to the radius AD. Then, because of the parallels BC, DF, it will be, as AD: AB :: DF : BC and :: AF : AC, which is the same as the theorem is in words. C A DB Note. The radius is equal, either to the sine of 90°, or the tangent of 45°; and is expressed by 1, in a table of natural sines, or by 10 in the log. sines. |