Page images

which it appears that the radius of any circle whatever* bears the same proportion to the sine in that circle of the arc which measures one of the acute angles of a right angled triangle, that the hypothenuse of the triangle does to the side opposite the acute angle.

It is customary, for conciseness, to represent the sides opposite the angles of a triangle by small letters of the same name with the large letters which are placed at the angles; which large letters are also employed as the algebraic representatives of the angles. Thus in the triangle above, a being the right angle, the hypothenuse opposite is expressed by a; the side ac opposite B is represented by b, and so for the other. The above proportions would, according to this method, be written thus

R: sin B::a:b

R: sin c::2:0 Both these proportions are expressed in the single rule printed in italics above.f

The two acute angles of a right angled triangle are together equal to a right angle or 90°. (Geom. B. 1, Prop. 25,

. . Cor. 4,) therefore they are complements of each other; hence sin C = COS B; and the second of the above proportions may be changed into

R: COS B: a :C

which may be translated into ordinary language thus; radius:

* It is important to observe that the same trigonometrical lines of angles or arcs containing the same number of degrees in two different circles bear the same relation to each other. Thus in the diagram above, cm: MP: CM: MP or,

(R : sin) of the smaller circ ::(R: sin) of the larger also,

MP: CP :: MC : CP or (R: cos) of the one ::(R: cos) of the other. + When R = 1, multiplying the second and third terms and dividing by the first, in the preceding proportions we have


= a sin e


cra sin c That is either side = the hypoth. X the sine of the angle opposite.


the cosine of one of the acute angles of a right angled triangle :: the hypothenuse : the side adjacent the acute angle.*

When any three terms of a proportion are given, the remaining term can be found. If the unknown term be one of the extremes, multiply the two means and divide by the other extreme; if the required term be a mean, multiply the two extremes and divide by the other mean.

The above proportions contain each of them two of the sides of a triangle, the sine or cosine of an angle, and radius. If the lengths of the sides be given in numbers, these numbers may be put in place of the small letters which represent the sides in the proportion, and the general form becomes so far adapted to a particular case in the solution of right angled triangles ; but if the angle is given in degrees, how are we to know its sine or cosine, for that is the quantity which enters into the proportion ; and how are we to know the numerical value of R? For the present the student must be satisfied with the reply, that he can find the numerical value of any trigonometrical line corresponding to an angle of any given number of degrees, in a table at the end of the work. This is TABLE III, entitled, A Table of Natural Sines, &c. The degrees for angles or arcs of every magnitude within the quadrant will be found at the top of the columns of the table, and the minutes in the first column on the left, if the given angle or arc be less than 45°; and the degrees will be found at the bottom of the page and the minutes on the right, if it be greater than 45°, the length of the sine or cosine will be found in the column under or over the degrees, as the case may be, and on the same horizontal line with the minutes. The title of the column must be looked for at top if the arc be less than 45°, and at bottom if the arc be greater. The other trigonometrical lines may be easily calculated from the sines and cosines, as will be seen in the examples.

* When R = 1

C = a COS B

or either side

hypoth. X cos adjac. angle.


The trigonometrical lines of this table are computed, by a rule which will be hereafter demonstrated, for a circle whose radius is 1. So far as the principles for the solution of triangles are concerned, the length of the radius is entirely immaterial, as it will be recollected that the arc in the diagram above was described with any radius at pleasure.

When in cases of the solution of right angled triangles, the hypothenuse and one of the acute angles are either given or required by the problem, one of the above formulæ* is always employed.

39. Let us take an example by which to illustrate their application, and as upon a former occasion, one which shall at the same time exhibit the practical utility of Trigonometry.

A roof is to have a height of 15 feet in the interior at the centre, and an inclination of 350.

Required the length of the inner line of the rafters.

A right angled triangle will be formed in which the angle at the base will be 35°, and the side opposite 15 feet, and of which the hypothenuse is required. The first formula of the last article applied to this case givest

1: sin 350 ::a: 15 Multiplying the extremes and dividing by the first mean, the value of the other mean which is a, the hypothenuse required, will be obtained

1 x 15 sin 350

[ocr errors]


a =

* They are under the form of a proportion, but may be converted into equations. This remark is here made because we have spoken of equations only as being called formulæ. Proportions are no less so.

+ Since in the demonsiration of the formulæ, the sides and angle of the triangle were supposed to have no particular values, it follows that any numbers, or any other letters compatible with the properties of a triangle, may be put in the place of those employed, and the formulæ will still be true. This must be borne in mind throughout the work.

a =

Looking out the sine of 35° in the tables and performing the operations indicated in the last equation, the value of a will be known which will be the length of the rafters required. The answer will be in feet. Sin 35° is found from the tables to be .57358, hence,

1 x 15

= 26.1 feet

.57358 If (to vary the problem) half the interior breadth of the roof had been given, say 20 feet, and the angle of inclination instead of 35o as in the last example had been 15°, then to find the length of the rafters, it would be necessary to find first the argle opposite the given side 20 feet; which is done by subtracting the given angle 15° from 90°, since the two acute angles are complements of each other. (Geom. B. 1, Prop. 25, Cor. 4.) The remainder is 750.

. Applying the same formula as before, there results the proportion

1: sin 750 ::a: 20 whence,

1 x 20

sin 750 The same result might be obtained by using the given angle 15°, and employing the last of the general formulæ above which contains the cosine of one of the acute angles. The proportion would stand thus

1: : cos 150 : :a: 20 whence,

1 x 20

[ocr errors]


a =

cos 15°

Finding cosine of 15° in the tables and substituting it here, the value of a would be found to be the same as in the last result. In fact cos 150 = sin 750.

40. Had the height and half the breadth of the interior of the roof been given, the length of the rafters might have been obtained, by employing the property of the right angled triangle demonstrated at Prop. 11, B. 4, of the Geometry, that the square on the hypothenuse is equivalent to the sum of the squares upon the other two sides. Let the height of the roof be 12 feet, and the semi-breadth 16 feet, then

12? + 162 400 whence,

a’ =

a = 20


If the length of the rafters had been given equal to 20 feet, and the height of the roof equal to 12 feet, then the semibreadth would have been expressed thus

b? = 20 - 122 256 whence,

b 16 41. Had the semi-breadth or base of the triangle and the inclination of the roof been given, and the height of the roof or perpendicular of the triangle been required, the hypothenuse not entering into the problem, neither of the above formulæ, all of which contain the hypothenuse, would serve to find the side required in a direct manner. It might, however, be found indirectly by first finding the hypothenuse, using one of the above proportions, and then by means of the hypothenuse, using the same proportion, the required side might be obtained.

It is, however, objectionable to find one of the required parts in terms of a part which has itself been calculated from the given parts; because in the use of the tables which give the trigonometrical lines of the different angles not with perfect accuracy, but truly for as many decimal places as the table employs, a small error arises from the decimals neglected beyond the last place, and this though so small as to be unimportant, becomes magnified by repetition, as in the case where one part of a triangle itself not perfectly accurate, is employed to calculate another. It is therefore desirable to find each of the required parts, in terms of the given parts; and this may always be done in right angled triangles. We proceed, therefore, to demonstrate a formula for the direct solution of the last case supposed above.

« PreviousContinue »