column, the base of which measures just 74 ft. to the centre of the figure's base. Required the distance between the tops of the two columns.

(40.) Two sides of a triangle are in the ratio of 11 to 9, and the opposite angles have the ratio of 3 to 1. What are these angles? (41.) The diagonals of a parallelogram are 12432 and 8413, and the angle between them is 78° 44′; find its area.

(42.) One side of a triangle is 1012.6 and two angles are 52° 21′ and 57° 32'; find its area.

(43.) Two sides of a triangle are 218.12 and 123.72, and the included angle is 59° 10'; find its area.

(44.) Two angles of a triangle are 35° 15′ and 47° 18′, and one side is 2104.7; find its area.

(45.) The three sides of a triangle are 1.2371, 1.4713, and 2.0721; find the area.

(46.) Two sides of a triangle are 168.12 and 179.21, and the included angle is 41° 14'; find its area.

(47.) The three sides of a triangle are 51 ft., 48.12 ft., and 32.2 ft.; find the area.

(48.) Two sides of a triangle are 111.18 and 121.21, and the included angle is 27° 50'; find its area.

(49.) The diagonals of a parallelogram are 37 and 51, and they form an angle of 65°; find its area.

(50.) If the diagonals of a quadrilateral are 34 and 56, and if they intersect at an angle of 67°, what is the area?

SPHERICAL TRIGONOMETRY

CHAPTER VIII

RIGHT AND QUADRANTAL TRIANGLES

RIGHT TRIANGLES

82. Let O be the centre of a sphere of unit radius, and ABC a right spherical triangle, right angled at A, formed by the intersection of the three planes AOC, AOB, and BOC

with the surface of the sphere. Suppose the planes DAC" and BEC' passed through the points A and B respectively, and perpendicular to the line OC. The plane angles DC"A and BC'E each measure the angle C of the spherical triangle, and the sides of the spherical triangle a, b, c have the same numerical measure as BOC, AOC, and AOB respec

Tan c

tively, then, AD=tanc, BE=sin c, BC' = sin a, OC' = cos a, OC" cos b, OE=cos c, AC" sin b.

=

In the two similar triangles OEC' and OAC",

Combining formulas (2) and (3) with (1),

Again, if AB were made the base of the right spherical

triangle ABC, we should have

NAPIER'S RULES OF CIRCULAR PARTS

83. The above ten formulas are sufficient to solve all cases of right spherical triangles. They may, however, be

expressed as two simple rules, called, after their inventor, Napier's rules.

The two sides adjacent to the right angle, the complement of the hypotenuse, and the complements of the oblique angles are called the circular parts.

The right angle is not one of the circular parts.

comp C

comp a

comp B

Thus there are five circular parts-namely, b, c, compa, comp B, compC Any one of the five parts may be called the middle part, then the two parts next to it are called adjacent parts, and the remaining two parts are called the opposite parts.

с

Thus if is taken for the middle part, comp B and b are adjacent parts, and compa and comp C are opposite parts.

The ten formulas may be written and grouped as follows:

I. The sine of the middle part is equal to the product of the tangents of the adjacent parts.

II. The sine of the middle part is equal to the product of the cosines of the opposite parts.

84. In the right spherical triangles considered in this work, each side is taken less than a semicircumference, and each angle less than two right angles.

In the solution of the triangles, it is to be observed,

(1.) If the two sides about the right angle are both less or both greater than 90°, the hypotenuse is less than 90°; if one side is less and the other greater than 90°, the hypotenuse is greater than 90°.

(2.) An angle and the side opposite are either both less or both greater than 90°.

or

EXAMPLE

85. Given a=63° 56′, b=40° 0', to find c, B, and C.