ered, take that one of the general equations for obliqueangled triangles, that will contain the three correspond ing parts of the triangle, together with the right angle; then make A = 90°, and after making the reductions corresponding to this supposition, the resulting equation will prove the rule for that particular case.

For example, let comp. a, be the middle part and the extremes opposite. The equation to be applied in this case must contain a, b, c, and A. The first of equations (1) contains these four quantities: cos a =

If A 90° cos A

hence,

cos b cos c + sin b sin c cos A.

that is, radius, which is 1, into the sine of the middle part, (which is the complement of a,) is equal to the rect angle of the cosines of the opposite parts.

Suppose, now, that the comple

that is, radius, which is 1, into the sine of the middle part is equal to the rectangle of the tangent of the complement of B, into the tangent of the complement of C, that is, to the rectangle of the tangents of the adjacent circular parts,

Let us now take the comp. B, for the middle part and the extremes opposite. The two other parts under consid eration will then be the perpendicular b and the comp. of the angle C. The equation to be applied must contain the four parts A, B, C, and b: it is the second of equations (2). cos B sin A sin C cos b cos A cos 0.

Making A = 90°, we have,

cos B sin C cos b.

Let comp. B be still the middle part and the extremes adjacent. The equation to be applied must then contain the four parts a, B, c, and A. It is similar to equa tions (10);

cot a sin c = cos c cos B + cot A sin B.

But, if A = 90°, cot A =

hence,

or,

0;

cot a sin c = cos c cos B:

cos B cot a tang c.

By pursuing the same method of demonstration when each circular part is made the middle part, and making the terms homogeneous, when we change the radius from 1 to R (Plane Trig., Art. 87), we obtain the five following equa tions, which embrace all the cases.

R cos a = cos b cos c = cot B cot C,

We see from these equations that, if the middle part is required we must begin the proportion with radius; and when one of the extremes is required we must begin the proportion with the other extreme.

We also conclude, from the first of the equations, that when the hypothenuse is less than 90°, the sides b and c are of the same species, and also that the angles B and C are likewise of the same species. When a is greater than 90°, the sides b and c are of different species, and the same is true of the angles B and C. We also see from the last two equations that a side and its opposite angle are always of the same species.

These properties are proved by considering the algebraic signs which have been attributed to the trigonometrical functions, and by remembering that the two members of an equation must always have the same algebraic sign.

SOLUTION OF RIGHT-ANGLED SPHERICAL TRIANGLES BY

LOGARITHMS.

16. It is to be observed, that when any part of a tri angle becomes known by means of its sine only, there may be two values for this part, and consequently two triangles that will satisfy the question; because, the same sine which corresponds to an angle or an arc, corresponds likewise to its supplement. This will not take place, when the unknown quantity is determined by means of its cosine, its tangent, or cotangent. In all these cases, the sign will enable us to decide whether the part in question is less or greater than 90°; the part is less than 90°, if its cosine, tangent, or cotangent, has the sign+; it is greater if one of these quantities has the sign

In order to discover the species of the required part of the triangle, we shall annex the minus sign to the logarithms of all the elements whose cosines, tangents, or cotangents, are negative. Then, by recollecting that the product of the two extremes has the same sign as that of the means, we can at once determine the sign which is to be given to the required element, and then its species will be known.

It has already been observed, that the tables are calculated to the radius R, whose logarithm is 10 (Plane Trig., Art. 100); hence, all expressions involving the circular functions, must be made homogeneous, to adapt them to the logarithmic formulas.

EXAMPLES.

1. In the right-angled spherical triangle BAC, right-angled at A, there are given a = 64° 40′ and b= 42° 12': required the remaining parts.

First, to find the side c.

B

a

The hypothenuse a corresponds to the middle part, and the extremes are opposite: hence,

R cos a = cos b cos c, or,

The side b is the middle part and the extremes oppo

site: hence,

R sin b = cos (comp. a) x cos (comp. B)

= sin a sin B.

B 48° 00' 14".

To find the angle C.

The angle C is the middle part and the extremes adja

2. In a right-angled triangle BAC, there are given the hypothenuse a = 105° 34', and the angle B = 80° 40′: re quired the remaining parts.

To find the angle C.

The hypothenuse is the middle part and the extremes adjacent: hence,

Since the cotangent of C is negative, the angle C is greater than 90°, and is the supplement of the arc which would correspond to the cotangent, if it were positive.

To find the side c.

The angle B corresponds to the middle part, and the extremes are adjacent: hence,

The side b is the middle part, and the extremes are

17. A quadrantal spherical triangle is one which has

one of its sides equal to 90°.

Let BAC be a quadrantal tri

angle of which the side a = 90°. If we pass to the corresponding polar triangle, we shall have

A

from which we see, that the polar triangle will be rightangled at A', and hence, every case may be referred to a right-angled triangle.

But we can solve the quadrantal triangle by means of the right-angled triangle in a manner still more simple.