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of the table, sine, tangent, &c. and against the minutes, is the sine, tangent &c. required. If the degrees are more than 45, look for them at the bottom of the table, and for the minutes, in the right hand column; then in the column marked at the bottom of the table, sine, tangent, &c. and against the minutes, is the sine, tangent, &c. required.
Note.-The sine' of an angle and of its supplement being the same, if the given number of degrees be above 90, subtract them from 180°, and find the sine of the remainder.
1. Required the sine of 32° 274 2. Required the tangent of 57° 39' 3. What is the secant of 89° 31'. 4. What is the sine of 157° 43'.
To find the degrees and minutes, corresponding to a gi
ven sine, tangent &c. Find, in the table, the nearest logarithm to the given one, and the degrees answering to it will be found at the top of the table if the name be there, and the minutes on the left hand; but if the name be at the bottom of the table, the degrees must be found at the bottom, and the minutes at the right hand.
1. Required the degrees and minutes in the angle whose sine is 9.64390. Ans. 26° 8.
2. Required the degrees and minutes in the angle whose tangent is 10.47464. Ans. 71° 28'.
ON GUNTER'S SCALE.
Gunters' Scale is an instrument by which, with a pair of dividers, the different cases in trigonometry and many other problems may be solved.
It has on one side, a diagonal scale, and also the lines of chords, sines, tangents and secants, with seyeral others.
On the other side there are several logarithmical lines as follow :
T'he line of numbers marked Num., is numbered from the left hand of the scale towards the right, with 1, 2, 3, 4, 5, 6, 7, 8, 9, 1 which stands in the middle of the scale; the numbers then go on 2, 3, 4, 5, 6, 7, 8, 9, 10 which stands at the right hand end of the scale. These two equal parts of the scale are similarly divided, the distances between the first 1, and the numbers 2, 3, 4, &c. being equal to the distances between the middle 1, and the numbers 2, 3, 4, &c. which follow it. The subdivisions of the two parts of this line are likewise simi. lar, each primary division being divided into ten parts, distinguished by lines of about half the length of the primary divisions.
The primary divisions on the second part of the scale, are estimated according to the value set upon the unit on
as a unit, then the first 1, 2, 3, &c. stand for 1, 2, 3, &c. the middle 1 is 10, and the 2, 3, 4, &c. following stand for 20, 30, 40, &c. and the 10 at the right hand for 100. If the first 1 stand for 10, the first 2, 3, 4, &c. must be counted 20, 30, 40, &c. the middle 1 will be 100, the second 2, 3, 4, &c. will stand for 200, 300, 400, &c. and the 10 at the right hand for 1000.
If the first 1 be considered as % of a unit, the 2, 3, 4, &c. following will be, i, o, &c. and the middle 1, and the 2, 3, 4, &c. following will stand for 1, 2, 3, 4, &c.
The intermediate small divisions must be estimated according to the value set upon the primary divisions.
Sines.-The line of sines, marked Sin. is numbered from the left hand of the scale towards the right, 1, 2, 3, 4, &c, to 10, then 20, 30, 40, &c to 90, where it terminates just opposite 10 on the line of numbers.
Tangents.-The line of tangents, marked Tan. begins at the left hand, and is numbered 1, 2, 3, &c. to 10, then 20, 30, 40, 45, where there is a brass pin, just under 90 in the line of sines; because the sine of 90° is equal to the tangent of 45°. From 45 it is numbered towards the left hand 50, 60, 70, 80, &c. The tangents of arcs above 45° are therefore counted backward on the line, and are found at the same points of the line as the tangents of their complements.
There are several other lines on this side of the scale, as Sine Rhumbs, Tangent Rhumbs, Versed Sines, &c.; but those described are sufficient for solving all the prob lems in plane trigonometry.
Remarks on Angles, Triangles, &c.
1. If from a point D in a right line AB, one or more right lines be drawn on the same side of it, the angles thus formed at the point D will be together equal to two right angles, or 180°; thus ADE + EDB = two right angles, or 180°: also ADC + CDE + EDB = two right angles, or 180°. Fig. 35.
2. Since the angles thus formed at the point D, on the other side of AB would also be equal to two right angles, the sum of all the angles formed about a point is equal to four right angles or 360.°
3. If two right lines cut one another, the opposite an. gles will be equal : thus AECEBED and AED=CEB.
4. The sum of the three angles of a plane triangle is equal to two right angles, or 180°.
5. If the sum of two angles of a triangle be subtracted from 180°, the remainder will be the third angle.
6. If one angle of a triangle be subtracted from 180°, the remainder will be the sum of the other two angles.
7. In right angled triangles, if one of the acute angles be subtracted from 90°, the remainder will be the other acute angle,
8. The angles at the base of an isosceles triangle are equal to one another.
9. If one side of a triangle be produced, the external posite angles : thus the external angle CBD, of the triangle ABC, is equal to the sum of the internal and opposite angles A and C. Fig. 37.
10. The angle at the centre of a circle is double of the angle at the circumference, upon the same base, that is, upon the same part of the circumference : thus the angle BEC is double of the angle BAC. Fig. 38.
11. The angles in the same segment of a circle are equal to one another : thus the angle BAD is equal to the angle BED ; also the angle BCD is equal to the angle BFD. Fig. 39.
12. The angle in a semicircle is a right angle; thus the angle ECF, Fig. 45, is a right angle.
13. This mark' placed on the sides or in the angles of a triangle, indicates that they are given ; and this marko placed in the same way, indicates that they are required.
PRACTICAL RULES FOR SOLVING ALL THE CASES
OF PLANE TRIGONOMETRY.
The angles and one side of any plane triangle being
given, to find the other sides.
As the sine of the angle opposite the given side,
* DEMONSTRATION. Let A BC, Fig. 40, be any plane triangle, take BF =AC, and upon A B let fall the perpendiculars CD and FE, which will be