AB

2. In the triangle DAB, DA90°, a = 112° 2′ 9′′, 67° 3' 14", to find the other parts.

Since in this example a is obtuse, DB is obtuse.

In the right-angled triangle ABC we have A 22° 2′ 9" and AB= 67° 3′ 14′′; let a be the middle part, then Ab, ac, will be adjacent parts, and we shall have

rad sin comp. A=tan b × tan comp. c; that is,

rad cos A tan b cot c.. tan b

D

and

C

B

rad cos A

cot c

a

therefore, the angle D 65° 27' 9".

Take now a for the middle part, then a and c will be opposite parts; hence

rad x sin a= cos comp. A X cos comp. c,

that is,

sin A sin c

rad sin a sin a sin c ... sin a =

rad

and a will be acute, because the opposite angle is acute

therefore BD110° 12' 44".

As we have now to find в, take a for the middle part, then

and B will be adjacent parts, therefore

125. Napier's rules for the solution of right angled spherical triangles, though applicable to all cases, do not give results of that degree of accuracy, which is sometimes required, when the sought part expressed by its sine is very small, or expressed by its cosine, is very near 90°. The following formulæ may in such cases be used.

I.

At Art. 86, by the formula which follows, (6), we have, R being 1,

1-cos p2 sin2 p

in a similar manner, we might obtain

changing p into a, and substituting the value of cos a, given

which is a formula to be employed, when в and c are given and a required.

II.

With the same data to find b use the formula

tan b = √{tan [}(в—c)+45°] tan [}(B+C)—45°] } derived from

The hypothenuse a and the side c being given to find the adjacent angle B, use the formula

derived in a manner similar to that in Case I,

Finally, to obtain b when the opposite angle в and the hypothenuse a are given, we have, by Napier's rules,

This last gives the value of b, x being calculated by the equation tan x = sin a sin B.

N. B. In the above five formulæ, R is supposed to be 1. To introduce it correctly when logarithms are applied to the formulæ, it is only necessary to observe that the two members of each must be homogeneous.

126. When, in the case considered at Art. 87, the only part required happens to be the side opposite the given angle, the finding of the other two angles then becomes merely a subsidiary operation, and the determination of the required side, by Napier's analogies, seems somewhat lengthy. But a shorter method of solution is deducible from the fundamental formula,

cos c = cos a cos b + sin a sin b cos c

...

(1).

For substituting cos a tan a for its equal sin a it becomes cos c = cos a (cos b + tan a sin b cos c),

Hence, to find the side c, we must determine a subsidiary angle o from the equation

1. In a spherical triangle are given a 38° 30', b = 70o,

Other formulas for the determination of c might be easily deduced from the same equation (1), but this is as short and as convenient as any. We might also introduce here a distinct formula for the determination of one of the angles A, by help of a subsidiary arc o; but as little or nothing would be gained, in point of brevity, over the process by Napier's analogies, we shall not stop to investigate it.

127. If where two angles and the included side are given, the angle opposite to the given side be the only part required, a more compendious method of solution may be obtained by introducing a subsidiary arc, as in the last case. Thus from the fundamental formula for the cosine of a side in terms of the three angles, might be obtained by aid of the polar triangles, a formula which becomes when cos a tan ▲ is substituted for sin a,

cos c = cos a (tan a sin в cos c

COS B);

Hence, having found a subsidiary angle w, by the equation coto tan A COS C . . (1);

the sought angle is determined by the equation