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APPLICATION OF TRIGONOMETRY TO THE NUMERICAL
COMPUTATION OF THE DIFFERENT PARTS OF OBLIQUE
That the different parts of an oblique angled plane triangle may be computed, two angles and a side, or two sides and an angle, must be given. Now when two of the angles are given, the third, which is the supplement of their sum, is known; hence, in that case, all the angles, and a side opposite to one of them are given, to determine the other two sides. When one of the angles is given, the supplement of it, which is the sum of the other two, is known; therefore when one of these other angles is determined, the remaining angle is obtained at once.
The formula which have been demonstrated in the theory of trigo'nometry furnish the following rules for the solution of the different
1. When two sides and an angle opposite to one of them, or two angles and a side opposite to one of them, are given.
As any side of a triangle is to the sine of its opposite angle, so is any other side of the triangle to the sine of its opposite angle; and as the sine of any angle of a triangle is to its opposite side, so is the sine of any other angle of the triangle to its opposite side.
Note. When in the first of the foregoing proportions the less of the given sides is opposite to the given angle, the angle determined from the proportion may be either acute or obtuse, and the problem solved is then said to be ambiguous. But as every triangle has two acute angles, which are those that are opposite the least two of the sides, when the greater of the given sides is opposite the given angle, the angle determined from the proportion must be acute, and therefore in that case the problem is not ambiguous.
2. When two sides and the contained angle are given.
If a perpendicular be demitted from the extremity of the shorter of the given sides, upon the greater, or the greater produced, the solution of the problem may be effected by computing the parts of the two right angled triangles, which will be formed by the perpendicular and the sides of the triangle. Or the angles opposite the given sides
be determined as follows. As the sum of the given sides is to their difference, so is the tangent of half the sum of their opposite angles to the tangent of half the difference of the same angles, (Trig. Prop. 6.)
And this half difference of the angles added to half their sum gives the angle opposite the greater side, and subtracted, leaves the angle opposite to the less side.
Then, the angles being known, the required side is obtained by the
3. When the three sides are given to find the angles.
As the base, or longest side, 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, made by a perpendicular from the opposite angle, (Trig. Prop.7.) Then the half difference added to half the base, gives the segment of it adjacent to the greater of the other two sides; and subtracted, leaves the segment of it adjacent to the less side.
The triangle is thus divided into two right angled triangles, in each of which the hypothenuse and one of the other sides are given ; whence the angles may be computed by the preceding rules.
But the angles may be otherwise determined from the sides by the following rules, deduced from the logarithmic formulæ, at page 83.
Rule 1. From half the sum of the three sides of the triangle subtract each side separately. To the sum of the logarithms of half the sum of the sides, and the three remainders, add 20-602060, and from half the sum subtract the sum of the logarithms of the sides including the required angle, and the remainder will be the logarithm sine of that angle.
Rule 2. From half the sum of the three sides of the triangle subtract the side opposite the required angle. Then to the logarithms of the half sum and remainder add 20, and from that sum subtract the sum of the logarithms of the sides including the required angle, and half the remainder will be the logarithm cosine of half the required angle.
Rule 3. From half the sum of the three sides of the triangle subtract each of the sides including the required angle; then to the logarithms of the two remainders add 20, and from the sum subtract the sum of the logarithms of the sides containing the angle, and half the remainder will be the logarithm sine of half the required angle.
On these rules it may be observed, that the first which gives the sine of the whole angle is ambiguous in itself; but the ambiguity may be removed, by the consideration that when the square of the side opposite the angle computed is greater than the sum of the squares of the other two sides, the angle must be obtuse; otherwise the angle is acute.
As a small arc can be found more correctly by even proportion from its sine than its cosine, and an arc near the termination of the quadrant can be found more correctly from its cosine than its sine, the former of the rules for determining half the angle of a triangle may be applied in preference, when the half angle to be computed exceeds half a right angle, and the latter when the required half angle is less than half a right angle.
In the triangle A B C, given A B 376, the angle A 48° 3' and the angle B 40° 14' to find the other parts.
The angle C being the supplement of the sum of A and B, is 91° 43'.
Given two sides of a triangle 654, and 460, and the angle opposite to the less of these sides 35° 12', to find the other parts.
Make BAC equal to the given angle, and AC equal to the longer of the given sides. With C as a centre, and the length
с of the shorter of the given sides as a radius, cut A D B in Band D; join CD,CB; then both the triangles ACB, ACD, correspond with the given data; this being what has been noticed as the ambiguous case in the
A DEB solution of oblique angled plane triangles.
CD being equal to C B, the angles CD B and C B D are equal ; hence the angle A CD, which is the difference of the angles C D B and C A B, is the difference of the anglès C B D and C A B; and the
Angle B 55° 2', supplement 124° 58' = angle A DC
90 14, supplement 89 46 = angle A CB
In the triangle A B C are given AB 848, AC 534, and the included angle A 31° 17', to find the other parts.
Let C D (see the first of the annexed figures) be a perpendicular from C upon AB, then in the right angled triangle ADC A D B A B D
And in the triangle A B C, as A D 456'4 ... 2.659309
: D B 391.6
2:592843 : : cot A 31° 177 10.216374
12.809217 : cot B 35° 18' .. .. 10149908
A 31° 17'
66 35, sup. 113° 25' ACB
Or the angles A C B and A B C may be found thus,
in 1382 3:140508
ACB + ABC
74° 21' 10:552859
Having the angles, the side B C may be found as above.
In the triangle A B C (see the first figure on the last page) are given A B 757, A C 586, and B C 649, to find the angles.
BD + AD and AD =
BD - AD