CASE IV. Given the hypothenuse and the base, to find the angles and the perpendicular.

r.c

SOLUTION. Cos A sine c== -; sec A=cosec c=

b

r.b

Having found the angles, the perpendicular may be found by Case I. or by Case II. OR, a= b22 = √ (b+c;.(b−c). CASE V. Given the hypothenuse and the perpendicular, to find the angles and the base.

Having found the angles, the base may be found by Case I. Or, c=√ b2 — a2=√ (b+a).(b−a).

CASE VI. Given the base and the perpendicular, to find the angles and the hypothenuse.

Or, having found the angles, the hypothenuse may be found by Case II. or by Case III.

SOLUTIONS OF THE DIFFERENT CASES OF OBLIQUE ANGLED PLANE TRIANGLES.

CASE I. Given the angles and one

side, to find the other two sides.

CASE II. Given two sides and an angle opposite to one of them, to find the other angles, and the third side.

sine B.a

b

sine A.b

||sin

sine c=

a

NOTE. If the given be obtuse; or if the side opposite to the given be the greater of the two given sides, then the sought is always acute in every other case it is ambiguous.

Two angles being now given, their sum deducted from 180°, leaves the third 4; hence the remaining side may be found by Case I.

CASE III. Given two sides and the angle contained between them, to find the other angles; and the third side.

SOLUTION. Tang (AB)= .cotc (X. 44.)

a+b

Then (AB)+(90°—c)= opposite to the greater side. And (90°c)-(AB) opposite to the less side.

OR THUS,

b:a::r:tang, and if a be the greater of the two given sides, will be greater than 45°.

r: -45°:: cot C: tang (ACB), (C. 109.)

The remaining side may be found from the angles, by Case I

CASE IV. Given the three sides to find the angles.

ВООК III.

SPHERICAL TRIGONOMETRY.

CHAPTER I.

DEFINITIONS, &C. OF SPHERICAL ANGLES, ARCS, AND TRI

ANGLES.

(A) SPHERICAL TRIGONOMETRY treats on the properties of spherical triangles, or the position and magnitudes of arcs of circles described upon the surface of a sphere or globe.

(B) A sphere, or globe, is a round solid body, each part of its surface being equidistant from its centre. This centre is common to every circle described on the surface of the sphere, wherein spherical trigonometry is concerned, and such circles are called great circles.

(C) All great circles divide the globe into two equal parts, and consequently bisect each other at the distance of a semi-. circle or 180°. Hence two arcs cannot enclose a space, unless they are both semi-circles.

(D) A spherical angle is the inclination of the planes of two great circles to each other, which circles intersect or meet each other on the surface of the sphere, in a point called the angular point. The inclination of these planes must always be measured on the arc of a great circle, 90° from the angular point.

(E) A spherical triangle is formed on the surface of the sphere by the intersection of three great circles, and consists of three sides and three angles; any three of which parts being given the rest may be found.

(F) All the sides of a spherical triangle are arcs of equal circles, and the angles of a spherical triangle are measured by arcs of circles, having the same radii as the sides. Hence the sides and angles of spherical triangles are always expressed in degrees, and parts of degrees.

(G) If one angle of a triangle be 90° degrees, it is called a right-angled triangle. If one side be 90°, a quadrantal triangle. If no angle or side be 90°, it is called an oblique-angled triangle.

(H) The

(H) The pole of a circle is a point on the surface of the sphere equidistant from every part of that circle of which it is the pole. Consequently every circle has two poles diametrically opposite to each other, and the arc of a circle comprehended between each of these poles, and the circumference of such a circle, is a quadrant. No two circles can have the same, or a common pole. If a straight line be drawn from the pole of any circle to the centre of the sphere, it will cut the diameter of that circle at right-angles.

(I) All great circles passing through the pole of another great circle, cut that circle at right-angles; and if two circles cut each other at right-angles, in the poles of a third circle, the four points of intersection with that third circle, will be the four poles of the cutting circles; viz. the two opposite points will be the poles of that circle which is described between them.

(K) Sides and angles of spherical triangles are said to be of the same species, kind, or affection, when by comparing any two sides, any two angles, or an angle and a side together, you discover each to be greater or less than a right-angle, or equal to a right-angle.

But when by comparing a side with a side, an angle with an angle, or a side with an angle, you discover one to be less and another greater than a right-angle; such sides and angles are said to be of different species.

(L) Spherical triangles are equilateral, isosceles, or scalene, according as they have three equal sides, two equal sides, or three unequal sides.

N.B. In any of the following propositions wherever the word circle, or arc of a circle occurs, it must always be understood to be a great circle, or the arc of a great circle. And, that all circles concerned in spherical trigonometry are equal to each other.

PROPOSITION I.

(M) If one great circle intersect another great circle in any point A, all the angles described from the point A, on the same side of any arc CAD, or EAB are equal to two rightangles; and all the angles made about the point A, are equal to four right-angles.

DEMONSTRATION. Let CA be perpendicular to EB, then will the angles EAC and CAB be each of them a right-angle. If AL be drawn, then LAB and EAL are equal to two right-angles, for the angle CAB is increased by the angle LAC, and CAE is diminished by the same, therefore LAB+EAL=EAC+CAB.

In the same manner it may be proved that the angles EAD and DAB are together equal to two right-angles, for if Ac be perpendicular to EB, AD will likewise be perpendicular to it, therefore all the angles about the point A are equal to four right-angles. Q.E.D.

(N) COROLLARY I. If two arcs of circles intersect each other, the vertical or opposite angles will be equal.

For the angles EAC and CAB are equal to two right-angles, also CAB and BAD are equal to two right-angles; from each of these equals take the angle CAB, then EAC is equal to BAD. In the same manner it may be shewn that the angle CAB is equal to the angle EAD.

(0) COROLLARY II. All the exterior and interior angles of a spherical triangle are together equal to six right-angles.

For the interior angle CAB and exterior angle BAD are equal to two right-angles; likewise ACB and BCG are equal to two right-angles; and ABC and CBG are also equal to two rightangles.

PROPOSITION II.

(P) If a great circle BC be described meeting two great circles ABG and ACG, which pass through the pole A of the circle BC; the angle CDB at the centre of the sphere, upon the circumference BC, is the

DT A

same with the spherical angle BAC, and the arc BC is called the sure of the spherical angle BAC.

DEMONSTRATION. Since A is the pole of BC, AB and AC are quadrants (H. 130.) and the angles ABC and ACB are rightangles (I. 130.)

Let D be the centre of the sphere, and join DC and DB; then because the arcs AC and AB are each of them quadrants, the angles ADC and ADB are right-angles (H. 130.): therefore the angle CDB is the inclination of the planes of the circles ABG and ACG to each other, and consequently, (D. 129.) it is the measure of the angle CAB. Q.E.D.

(Q) COROLLARY. If two circles cut each other in two points, the angles at these points are equal to each other, and to the distance between the poles of these circles. Viz. the angle BAC is equal to the angle BGC, for the arc Bc is the measure of them both; and the distance between the point B and the pole of ABG is a quadrant, also the distance between c and the pole of ACG is a quadrant, (H. 130.) therefore the difference of their distances will be BC.