8. In a right triangle is given

c = 75° 25',

a = 52° 16';

find the lengths of the segments into which a is divided by the bisector of A.

EXERCISES IN GEOMETRIC APPLICATION.

9. From a point P above a plane an oblique line PO is drawn, meeting the plane in O and making the angle A with the plane. Let be the projection of P upon the plane, so that OQ is the projection of OP. Through O a line OM is drawn, making an angle QOM = B with the projection OQ. It is required to express the angle POM in terms of A and B.

Ans. Cos POM = cos A cos B.

10. In the preceding case, if a perpendicular PS be dropped from P upon OM, express the length OS in terms of the angle A and B and the length OP. Ans. OS OP cos A cos B.

=

11. Two planes intersect at right angles along a line I. From any point R of I one line is drawn in each plane, making the respective angles A and B with I. Express the angle С between these lines. Ans. Cos C = cos A cos B.

12. Two planes intersecting at right angles along a line I are intersected by a third plane, making with them the respective angles P and Q. Express the angles which the three lines of intersection make with each other.

Ans. If we put PI for the angle between I and that edge along which the dihedral angle P is formed, etc., we have

116. Isosceles triangles. An isosceles spherical triangle may be divided into two symmetrical right triangles by a perpendicular from its vertex upon its base. If we put

c, each of the equal sides;

C, each of the equal angles at the base;

b, the base, or third side;

B, the angle at the vertex;

P, the middle point of b;

p, the length of the perpendicular BP from B upon b,we shall then have two right triangles in each of which the oblique angles are C and B, the hypothenuse is c, and the sides containing the right angle are p and b. The equations of § 109 will then give

1. The equal sides of an isosceles triangle are each 45°, and the angle which they contain is 95°. Find the base and the angles at the base.

2. If the base of an isosceles triangle is 95°, and the angles at the base each 45°, find the remaining parts.

117. Quadrantal triangles. Since the polar of a right triangle is a quadrantal triangle, the formula for quadrantal triangles may be obtained by applying the formulæ of § 109 to the polar triangle. The side e will then be a quadrant, and the relations among the other parts will be

If we take, as the five parts of the triangle,

A, B, 90° a, 90°-b, C-90°,

(a)

and omit the hypothenuse c, the above formulæ will be expressed by a set of rules identical in expression with those of Napier. For example, let us consider the parts a, b, C. Here C will be a

middle part, and a and b adjacent parts. Applying Napier's rules to this case, with the parts (a) we have

sin (C-90°): =tan (90° - a) tan (90° — b);

an equation identical with the fifth of the above list.

EXERCISES.

1. Let the student deduce the six equations (15) by applying Napier's rules to the parts (a).

2. Through the same point there pass two lines intersecting at right angles, and a plane P making the angle a with one of the lines, and the angle ẞ with the other. Express the angle which the plane P forms with the plane of the lines.

Ans. Sin A = √sin2 a + sin2 ß.

3. The sides of an obelisk have a slope of 8° from the perpendicular. What is the face-angle at the base of the obelisk, the slope of the edges, and the dihedral angle between two adjacent lateral faces? Ans. Face-angle at base, 82° 4'.6; slope, 11° 14'.5; dihedral angle, 91° 6'.6.

In this problem, to reduce to a spherical triangle, consider the centre of the sphere to be at a corner of the obelisk. The slope of the edge will not be represented by either of the six parts of the triangle, but by the complement of the perpendicular from the vertex upon the base.

4. A mason cuts a stone with a rectangular base and four lateral edges, each making an angle of 60° with the base at its corners. What is the inclination of each lateral face to the base, and the dihedral angle between the faces, supposing such inclinations and dihedral angles all equal? Ans. 67° 47'.5 and 98° 12′.8.

5. In another stone the base is rectangular; one lateral face makes an angle of 68° 29' with the base, and the lateral edge bounding this face makes an angle of 52° 15' with the base. What angles does the adjacent lateral face make with the first face and with the base?

6. When the angular distance of the sun from the south point of the horizon is 75°, and from the west point 60°, what is its altitude above the horizon?

CHAPTER III.

TRANSFORMATION OF THE FORMULE OF SPHERICAL
TRIGONOMETRY.

118. Although the formulæ already given suffice for the solution of every spherical triangle, there are many transformations which will facilitate the applications of spherical trigonometry, and render the solutions of triangles more accurate and convenient. Let us first take the fundamental equation (1) of Chapter I., cos a = cos b. cos c + sin b sin c cos A.

(§ 43)

(2)

2 sin b sin c = cos (b+c) cos (b → c),

we have, by substituting,

L

cos a = cos (b— c) (1 sin' 4)+cos (b+c) sin2 A

= cos (b — c) cos2 4A + cos (b + c) sin2 †A.

This last equation may also be derived by the following elegant process. The original fundamental equation may be written

A and of sin2 4, the

By conjoining the coefficients of cos' equation (2) follows by the addition theorem.

By a similar process, from the equation

we obtain

cos A = - cos B cos C+ sin B sin C cos a,

cos A cos (B+ C) cos Acos (B+ C) cos2 By a slight modification of the the equations (1) and (3) we may find

sin B sin C sin' ta; (3) a · -cos (BC) sin' a. (4) sin❜a. process employed in forming

cos a = cos (b+c) + 2 sin b sin c cos2 A;

cos Acos (BC) + 2 sin B sin C cos' ta; which equations the student may prove as an exercise.

(5) (6)

119. Expressions when three sides or three angles are given. From the last equation (1) we find

by which any angle is expressed in terms of the three sides.

By § 44, 13 we have

cos (bc) - cos a = 2 sin (a+c-b) sin (a+b−c). (a) If we put s for half the sum of the sides, namely,

Substituting these values in (a), the expression for sin' A

To find similar expressions for the cosines we take equation (5), which gives

cos (b+c)

2 sin b sin c

cos acos(b+c) = 2 sin (b+c+a) sin (b+c-a)