Part 25

In finding the cosines of small angles it will be seen that direct reading on the rule becomes impossible for angles of less than 20°. It is advisable in such cases to adopt the method described for determining the sines of the large angles of which the complements are sought.

Cotangents of Angles.—From the methods of finding the tangents of angles previously described, it will be apparent that the cotangents of angles may also be obtained with equal facility. For angles between 5° 45′ and 45°, the procedure is the same as that for finding tangents of angles greater than 45°. Thus, the angle on scale T is brought to the R.H. index of D, and the cotangent read off on D under the L.H. index of T. The first figure of the result so found is to be read as an integer.

If the angle (θ) lies between 45° and 84° 15′, the slide is placed so that the indices of T coincide with those of D, and the result is then read off on D opposite (90 − θ) on T. In this case the value is wholly decimal.

Secants of Angles.—The secants of angles are readily found by bringing (90 − θ) on S to the R.H. index of A and reading the result on A over the L.H. index of S. If the value is found on the L.H. scale of A, the first figure is to be read as an integer; while if the result is read on the R.H. scale of A, the first two figures are to be regarded as integers.

Cosecants of Angles.—The cosecants of angles are found by placing the angle on S to the R.H. index of A, and reading the value found on A over the L.H. index of S. If the result is read on the L.H. scale of A, the first figure is to be read as an integer; while if the result is found on the R.H. scale of A, the first two figures are to be read as integers.

It will be noted that some of the rules here given for determining the several trigonometrical functions of angles apply only to those forms of rules in which a single scale of tangents T is used, reading from left to right. For the other arrangements of the scale, previously referred to, some slight modification of the method of procedure in finding the tangents and cotangents of angles will be necessary; but as in each case the nature and extent of this modification is evident, no further directions are required.

THE SOLUTION OF RIGHT-ANGLED TRIANGLES.

From the foregoing explanation of the manner of determining the trigonometrical functions of angles, the methods of solving right-angled triangles will be readily perceived, and only a few examples need therefore be given.

Let a and b represent the sides and c the hypothenuse of a right-angled triangle, and a° and b° the angles opposite to the sides. Then of the possible cases we will take

(1.) Given c and a°, to find a, b, and b°.

The angle b° = 90 − a°, while a = c sin a° and b = c sin b°. To find a, therefore, the index of S is set to c on A, and the value of a read on A opposite a° on S. In the same manner the value of b is obtained.

EX.—Given in a right-angled triangle c = 9 ft. and a° = 30°. Find a, b, and b°.

The angle b° = 90 − 30 = 60°. To find a, set R.H. index of S to 9 on A, and over 30° on S read a = 4·5 ft. on A. Also, with the slide in the same position, read b = 7·8 ft. [7·794] on A over 60° on S.

(2.) Given a and c, to determine a°, b°, and b.

In this case advantage is taken of the fact that in every triangle the sides are proportional to the sines of the opposite angles. Therefore, as in this case the hypothenuse c subtends a right angle, of which the sine = 1, the R.H. index (or 90°) on S is set to the length of c on A, when under a on A is found a° on S. Hence b° and b may be determined.

(3.) Given a and a°, to find b, c, and b°.

Here b° = (90 − a°), and the solution is similar to the foregoing.

(4.) Given a and b, to find a°, b°, and c.

To find a°, we have tan. a° = a/b, which in the above example will be (4·5)/(7·8) = 0·577. Therefore, placing the slide so that the indices of T coincide with those of D, we read opposite 0·577 on D the value of a° = 30°. The hypothenuse c is readily obtained from c = a/(sin a°).

THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES.

Using the same letters as before to designate the three sides and the subtending angles of oblique-angled triangles, we have the following cases:—

(1.) Given one side and two angles, as a, a°, and b°, to find b, c, and c°.

In the first place, c° = 180° − (a° + b°); also we note that, as the sides are proportional to the sines of the opposite angles, b = (a sine b°)/(sine a°) and c = (a sine c°)/(sine a°).

Taking as an example, a = 45, a° = 57°, and b° = 63°, we have c° = 180 − (57 + 63) = 60°. To find b and c, set a° on S to a on A, and read off on A above 63° and 60° the values of b (= 47·8) and c (= 46·4) respectively.

(2.) Given a, b, and a°, to find b°, c°, and c.

In this case the angle a° on S is placed under the length of side a on A and under b on A is found the angle b° on S. The angle c° = 180 − (a° + b°), whence the length c can be read off on A over c° on S.

(3.) Given the sides and the included angle, to find the other side and the remaining angles.

If, for example, there are given a = 65, b = 42, and the included angle c° = 55°, we have (a + b) ∶ (a − b) = tan. (a° + b°)/(2) ∶ tan. (a° − b°)/(2). Then, since a° + b° = 180° − 55° = 125°, it follows that (a° + b°)/(2) = (125°)/(2) = 62° 30′.