Again, in the triangle BPC, right-angled at P,

So that all the three angles are as follow, viz.

the LA 27° 4' ; the B 37° 20'; the Lc 115° 36'.

The angles A and B may also easily be found by the ex

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 tri angles (see vol. ii.), 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 explained.

THEOREM IV.

When a Triangle is Right-angled; any of the unknown parts may be found by the following proportions: viz.

As radius

Is to either leg of the triangle;

So is tang, of its adjacent angle,

To its opposite leg;

And so is secant of the same angle,
To the hypothenuse.

E

F

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 BC. 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

D B

as AD: AB:: DF: BC and :: AF: AC, which is the same as the theorem is in words.

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.

Make AB

the leg AB 162

LA 53° 7' 48"

To find AC and BC.

1. Geometrically.

162 equal parts, and the angle A=53° 7′ 48°;

then raise the perpendicular BC, meeting AC in c. So shall

AC measure 270, and BC 216.

Note. There is sometimes given another method for rightangled triangles, which is this:

ABC being such a triangle, make one leg AB radius; that is, with centre A, and distance AB, describe an arc BF. Then it is evident that the other leg BC represents the tangent, and the hypothenuse AC the secant, of the arc BF, or of the angle A.

F

B D

E

In like manner, if the leg Bc be made radius ; then the other leg AB will represent the tangent, and the hypothenuse AC the secant, of the arc BG or angle c.

But if the hypothenuse be made radius; then each leg will represent the sine of its opposite angle; namely, the leg AB the sine of the arc AE or angle c, and the leg BC the sine of the arc CD or angle a.

Then the general rule for all these cases is this, namely, that the sides of the triangle bear to each other the same proportion as the parts which they represent.

And this is called, Making every side radius.

Note 2. When there are given two sides of a right-angled triangle, to find the third side; this is to be found by the property of the squares of the sides, in theorem 34, Geom. viz. that the square of the hypothenuse, or longest side, is equal to both the squares of the two other sides together. Therefore, to find the longest side, add the squares of the two shorter sides together, and extract the square root of that sum; but to find one of the shorter sides, subtract the one square from the other, and extract the root of the remainder. Or, when the hypothenuse, H, and either the base, B, or the perpendicular, P, are given: then half the sum of log. (H+P) and log. (n log. B; and half the sum of log. (H+B) and log. (HB) = log. P.

=

When в and P are given, the following logarithmic operation may sometimes be advantageously employed; viz. Find N the number answering to the log. diff., 2 log. P — log. B; and make в + NM: then, (log. м + log. B) = log. н, the hypothenuse.

The truth of this rule is evident: for, from the nature p2

of logarithms.

p2

B

=N; whence в + NB +

B)

B

=м; and (log. м + log. в) = log. MB = log.

=

(B3 + p2) = log. √ (B2 + P2) log. H.

Or, still more simply, find 10+ the diff. (log. rlog. B) in the log. tangents. The corresponding log. secant added to log. B= log. H.

Note, also, as many right-angled triangles in integer numbers as we please may be found by making

hypothenuse

=

m2 n2 perpendicular

--

m and n being taken at pleasure, m greater than n.

Before we proceed to the subject of Heights and Distances we shall give,

A CONCISE INVESTIGATION OF SOME OF THE MOST USEFUL TRIGONOMETRICAL FORMULE.

also, because AC = (AB+ AD) = centre) = BDQ (at circumf.)

{BAQ angle Aoc (at BMQ (on samne arc) .. triangles AOH, вDK, QмK, are equiangular.

Hence

:

I. OA: AH :: MQ : QK ;

that is, rad. sin. AC: : 2 cos. BC: sin. AD + sin. ae II. AO: OH :: BD : DK;

or, rad. cos. AC :: 2 sin. BC: sin. AD

III. Aо: OH: QM: MK;

sin. AB

or, rad. cos. AC: : 2 cos. BC: Cos. AB+ COS. ad
IV. AO: AH: : DB: DK;

or,

rad. : sin. AC 2 sin. BC: cos. AB COS AD;

also,

V. BK. KM = DK. KQ, that is (cos. Ab — cos. Ad) (cos. AB+COS. AD) = (sin. AD-sin. AB) (sin. AD+sin. AB).

By reducing the above four proportions into equations, making rad. 1, we obtain two distinct classes of formulæ,

thus:

First Class. AC = a, CB = b; then AD = a + b, AB = a~b,

(a + b),

Second Class. AD α, AB = b; then Ac=

BC= α- - b).

5. sin. a + sin. b = 2 sin.
6. sin. a sin. b
7. cos.b+cos. a 2 cos.
cos. a= 2 sin.

8. cos.b

(a + b) sin. 1(a - b)

The first class is useful in transforming the products of sines into simple sines, and the contrary.

The second facilitates the substitution of sums or differences of sines for the products, and the contrary.

Taking the sum and the difference of equations 1 and 2, also of 3 and 4, remembering that sin. cos. tan. we obtain the following:

9. sin. (a + b)

Third Class.

sin. a cos. b + sin. b cos. a =cos. a cos. b (tan. a + tan. b)