to the common method, though I have thought it right Trigonom to point out that which should in strictness be pursued

A proposition is left out in the Plane Trigonometry, which the astronomers make use of, in order, when two sides of a triangle, and the angle contained by them, are given, to find the angles at the base, without making use of the sum or difference of the sides, which, in some cases, when only the Logarithms of the sides are given, cannot be conveniently found.

THEOREM.

.:

If, as the greater of any two sides of a triangle to the less, so the radius to the tangent of a certain angle; then will the radius be to the tangent of the difference between that angle and half a right angle, as the tangent of half the sum of the angles at the base of the triangle to the tangent of half their difference.

Let ABC be a triangle, the sides of which are BC and CA, and the base AB, and

let BC be greater than CA.

Let DC be drawn at right IV 103

angles to BC, and equal to AC; join BD, and because

(Prop. 1.) in the right

C

angled triangle BCD, BC: Duri anaitisoque Bea

CD:: R: tan CBD, CBD is the angle, of which the tangent is to the radius as

CD to BC, that is, as CA to BC, or as the least of two sides of the triangle to the greatest.

But BC + CD: BC-CD :: tan (CDB + CBD) :

tan (CDB-CBD) (Prop. 5.):

(CAB+CBA);

and also, BC + CA: BC-CA:: tan
tan (CAB-CBA). Therefore, since CD=CA,
tan (CDB + CBD) : tan † (CDB-CBD): ;

}

Trigonom. tan (CAB+CBA): tan (CAB-CBA). But because the angles CDB+CBD=90°, tan & (CDB+CBD): tan (CDB-CBD)::R:tan (45°-CBD), (2 Cor. Prop.3.); therefore, R: tan (45°-CBD) :: tan ! (CAB+CBA): tan 1 (CAB-CBA); and CBD was already shewn to be such an angle that BC: CA::R: tan CBD. Therefore, &c. Q. E. D.

COR. If BC, CA, and the angle C are given to find the angles A and B; find an angle E such, that BC: CA

R: tan E; then R: tan (45°-E) :: tan ! (A+B): : tan (AB). Thus (A-B) is found, and I (A+B) being given, A and B are each of them known, Lem. 2

In reading the elements of Plane Trigonometry, it may be of use to observe, that the first five propositions contain all the rules absolutely necessary for solving the different cases of plane triangles. The learner, when he studies Trigonometry for the first time, may satisfy himself with these propositions, but should by no means neglect the others in a subsequent perusal.

I have changed the demonstration which I gave ofs these propositions in the first edition, for two others considerably simpler and more concise, communicated to me by Mr JAMES JARDINE, Civil Engineer, well known for his ingenuity and skill, both in the pure and mixed mathematics.

T

THE following theorem is given, as being of great utility in the construction and use of mathematical instruments, and particularly as containing one of the main principles of the modern improvement which has substituted the entire circle for the quadrant, the semicircle, and other segments formerly used in the instruments of Astronomy and Practical Geometry.

Though the proposition is quite elementary, being in reality an extension of the 20th of the 3d of Euclid, it has been reserved to this place, as being best enunciated in the language of Trigonometry.

*

If two straight lines intersect one another in a point within a circle, half the sum of the op posite arcs cut off by them is the measure of the angle which they contain.

Let BE and CD cut one another in the point A: half the sum of the arcs BC, DE is the measure of the angle BAC.

Through E, draw EF parallel to DC, meeting the circumference in F; the arc CF is therefore equal to

Trigonom. DE. Add BC to both, and the arc BF is equal to the sum of the arcs BC and DE. But because the angle BEF at the circumference, stands on the arc BF, it is equal to half the angle at the centre standing on the same arc, and is therefore measured by half that arc. But the angle BEF is equal to the angle BAC, therefore the angle BAC is measured by half the arc BF, that is by BC + DE or by half the sum of the opposite arcs, which

2

AB and CD intercept.

COB. 1. It is demonstrated in the same way, that if the point in which the lines intersect be without the circle, half the difference of the arcs which they intercept is the measure of the angle which they contain.

COR. 2. It is evident, that if the circumference of a circle be divided into degrees, and parts of a degree, half the sum of the opposite arcs cut off by any two lines will give the degrees and parts of a degree that measure the angle made by the two lines, whether they intersect one another in the centre or not,

2

PROP. V. SPH, TRIG.

The angles at the base of the isosceles spherical triangle are Symmetrical magnitudes, not admitting of being laid on one another, nor of coinciding, notwithstanding their equality. It might be considered as a suf ficient proof that they are equal, to observe that they are each determined to be of a certain magnitude rather than any other, by conditions which are precisely the same, so that there is no reason why one of them should be greater than another. For the sake of those to whom this reasoning may not prove satisfactory, the demonstration in the text is given, which is strictly geometrical.

Trigonom.