73. If A = a, show that B and b are either equal or supplemental, as also C and c.

that a 2 AD.

B+C, and D be the middle point of a, show

75: When does the polar triangle coincide with the primitive triangle?

76. If D be the middle point of

C,

show that

78. If b+c=π, show that sin 2B + sin 2 C = 0.

80. If D be any point in the side BC of a triangle, show that

cos AD sin a = cos c sin DC + cos b sin BD.

C

81. Prove cos2=cos2+(a+b) sin2+cos2 + ( a − b ) cos2

=

2

2

sinsin (a+b) sine C,

2+sin2 + (a−b) cos2
+sin(a-b)

sin s sin (sa) sin (s — b) sin (s — c)

2

= (1 — cos2 a — cos2 b - cos2 c + 2 cos a cos b cos c).

84. If AD be the bisector of the angle A, prove that

85. Prove cos a sin b

=

sin a cos b cos C + cos A sin c.

86. 66

sin C cos a = cos A sin B + sin A cos B cos C.

89. If 8 be the length of the arc from the vertex of an isosceles triangle, dividing the base into segments & and ß,

91. If AB, AC be produced to B', C', so that BB', CC' shall be the semi-supplements of AB, AC respectively, prove that the arc B'C' will subtend an angle at the centre of the sphere equal to the angle between the chords of AB, AC.

CHAPTER XI.

SOLUTION OF SPHERICAL TRIANGLES.

199. Preliminary Observations. In every spherical triangle there are six elements, the three sides and the three angles, besides the radius of the sphere, which is supposed constant. The solution of spherical triangles is the process by which, when the values of any three elements are given, we calculate the values of the remaining three (Art. 184, Note).

In making the calculations, attention must be paid to the algebraic signs of the functions. When angles greater than 90° occur in calculation, we replace them by their supplements; and if the functions of such angles be either cosine, tangent, cotangent, or secant, we take account of the change of sign.

It is necessary to avoid the calculation of very small angles by their cosines, or of angles near 90° by their sines, for their tabular differences vary too slowly (Art. 81). It is better to determine such angles, for example, by means of their tangents.

We shall begin with the right triangle; here two elements, in addition to the right angle, will be supposed known.

SOLUTION OF RIGHT SPHERICAL TRIANGLES.

200. The Solution of Right Spherical Triangles presents Six Cases, which may be solved by the formulæ of Art. 185. If the formula required for any case be not remembered, it is always easy to find it by Napier's Rules (Art.

186).

In applying these rules, we must choose the middle part as follows:

When the three parts considered are all adjacent, the one between is, of course, the middle part. When only two are adjacent, the other one is the middle part.

Let ABC be a spherical triangle, right-angled at C, and let a, b, c denote the sides opposite the angles A, B, C, respectively.

We shall assume that the parts are all positive and less than 180° (Art. 182).

A

201. Case I. Given the hypotenuse c and an angle A; to find a, b, B.

By (3), (5), and (8) of Art. 185, or by Napier's Rules, we have

sin a = sin c sin A,

tan b = tan c cos A,

cot B = cos c tan A.

Since a is found by its sine, it would be ambiguous, but the ambiguity is removed because a and A are of the same species [Art. 187, (1)]. B and b are determined immediately without ambiguity.

If a be very near 90°, we commence by calculating the values of b and B, and then determine a by either of the formulæ

tan a sin b tan A, tan a = tan c cos B.

Check. As a final step, in order to guard against numerical errors, it is often expedient to check the logarithmic work, which may be done in every case without the necessity of new logarithms. To check the work, we make up a formula between the three required parts, and see whether

it is satisfied by the results. In the present case, when the three parts a, b, B have been found, the check formula is

Ex. 1. Given c = 81° 29′ 32", A = 32° 18′ 17′′; find a, b, B.

A = 80° 10' 30"; find a, b, B.

Ex. 2. Given c = 110° 46' 20",

Ans. a = 67° 5' 52".7, b = 155° 46′ 42′′.7, B = 153° 58′ 24′′.5.

202. Case II. Given the hypotenuse c and a side a; to find b, A, B.

By (1), (3), (4) of Art. 185, or by Napier's Rules, we have

The check formula involves b, A, B; therefore, from (9) of Art. 185 we have

In this case there is an apparent ambiguity in the value of A, but this is removed by considering that A and a are always of the same species (Art. 187).