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and A are each right angles, and then b and B become indeterminate.

It is easy to see from a figure that the ambiguity must occur in general.

B

A.

C

For, suppose BAC to be a triangle which satisfies the given conditions ; produce AB and AC to meet again at A'; then the triangle A'BC also satisfies the given conditions, for it has a right angle at C, BC the given side, and A' = A the given angle.

If a=A, then the formulæ of solution shew that c, b, and B are right angles; in this case A is the pole of BC, and the triangle

: A'BC is symmetrically equal to the triangle ABC (Art. 57).

If a and A are both right angles, B is the pole of AC; B and 6 are then equal, but may have any value whatever.

There are limitations of the data in order to insure a possible triangle. A and a must have the same affection by Art. 64; hence the formulæ of solution shew that a must be less than A if both are acute, and greater than A if both are obtuse.

a

EXAMPLES.

a

If ABC be a triangle in which the angle C is a right angle, prove the following relations contained in Examples 1 to 5.

b

b 1. Sin = sino q cose + cosa sin ?

2

6 2. Tan 3(e+a) tan 1(c-a) = tan

2

2

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2

2

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4. Sin a tan / A – sin b tan 1 B= sin (a - b). 5.

Sin (c - a)= sin bcos a tan 1B,

Sin (c- a)= tan b cos c tan 1 B. 6. If ABC be a spherical triangle, right-angled at C, and cos A = cos a, shew that if A be not a right angle b+c=or 3 , according as b and e are both less or both greater than

2

T

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7. If a, ß be the arcs drawn from the right angle respectively perpendicular to and bisecting the hypotenuse c, shew that

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8. In a triangle, if C be a right angle and D the middle point of AB, shew that

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ز

n

1

2

9. In a right-angled triangle, if : be the length of the arc drawn from C perpendicular to the hypotenuse AB, shew that

cot d = (cota+ cot b). 10. OAA, is a spherical triangle right-angled at A, and acuteangled at A ; the arc 4, 4, of a great circle is drawn perpendicular to OA, then 4,4, is drawn perpendicular to 04,, and so on : shew that A A+, vanishes when n becomes infinite; and find the value of cos AA, cos 4, A, cos 4,4 3......to infinity.

11. ABC is a right-angled spherical triangle, A not being the right angle : shew that if A = a, then c and 6 are quadrants.

12. If 8 be the length of the arc drawn from C perpendicular to AB in any triangle, shew that

= cosec c (cosoa + cos® 6 – 2 cos a cos b cos c)). 13. ABC is a great circle of a sphere; AA', BB', CC', are arcs of great circles drawn at right angles to ABC and reckoned posi

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cos 8

2

tive when they lie on the same side of it: shew that the condition of A, B, C lying in a great circle is

tan AA' sin BC + tan BB' sin CA + tan CC" sin AB=0.

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14. Perpendiculars are drawn from the angles A, B, C of any triangle meeting the opposite sides at D, E, F respectively: shew that

tan BD tan CE tan AF=tan DC tan EA tan FB.

=

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15. Ox, Oy are two great circles of a sphere at right angles to each other, P is any point in AB another great circle. OC =p is the arc perpendicular to AB from 0, making the angle COx= a with Ox. PM, PN are arcs perpendicular to Ox, Oy respectively: shew that if OM = ? and ON = Y,

cos a tan x + sin a tan y= tan p.

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16. The position of a point on a sphere, with reference to two great circles at right angles to each other as axes, is determined by the portions 0, $ of these circles cut off by great circles through the point, and through two points on the axes, each from their

2 point of intersection : shew that if the three points (0, 4), (0', $'), (0", $'') lie on the same great circle

tan • (tan 6' – tan 6'') + tan $' (tan 6" – tan )

+ tan $" (tan 6 – tan 6') = 0. 17. If a point on a sphere be referred to two great circles at right angles to each other as axes, by means of the portions of these axes cut off by great circles drawn through the point and two points on the axes each 90° from their intersection, shew that the equation to a great circle is

tan 0 cot a + tan cot ß = 1.

=

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VI. SOLUTION OF OBLIQUE-ANGLED TRIANGLES.

78. The solution of oblique-angled triangles may be made in some cases to depend immediately on the solution of right-angled triangles; we will indicate these cases before considering the subject generally.

(1) Suppose a triangle to have one of its given sides equal to a quadrant. In this case the polar triangle has its corresponding angle a right angle; the polar triangle can therefore be solved by the rules of the preceding Chapter, and thus the elements of the primitive triangle become known.

(2) Suppose among the given elements of a triangle there are two equal sides or two equal angles. By drawing an arc from the vertex to the middle point of the base, the triangle is divided into two equal right-angled triangles ; by the solution of one of these right-angled triangles the required elements can be found.

(3) Suppose among the given elements of a triangle there are two sides, one of which is the supplement of the other, or two angles, one of which is the supplement of the other. Suppose, for example, that b+c=n, or else that B + C =n; produce BA and BC to meet at B' (see the first figure to Art. 38); then the triangle B'AC has two equal sides given, or else two equal angles given; and by the preceding case the solution of it can be made to depend on the solution of a right-angled triangle.

79. We now proceed to the solution of oblique-angled triangles in general. There will be six cases to consider.

80. Having given the three sides.

cos 6 cos c Here we have cos A

and similar formulæ

sin b sin c for cos B and cos C. Or if we wish to use formula suited to loga

COS a =

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rithms, we may take the formula for the sine, cosine, or tangent of half an angle given in Art. 45. In selecting a formula, attention should be paid to the remarks in Plane Trigonometry, Chap. XII. towards the end.

81. Having given the three angles.

cos A + cos B cos C Here we have cos a=

and similar formula

sin B sin c for cos b and cos C. Or if we wish to use formulæ suited to logarithms, we may take the formula for the sine, cosine, or tangent of half a side given in Art. 49.

a

There is no ambiguity in the two preceding cases; the triangles however

may be impossible with the given elements.

i in

82. Having given two sides and the included angle (a, C, b). By Napier's analogies

cos 1 (a - b) tan 3 (A + B)

cot 1C, cos } (a+b)

sin } (a-6) tan} (A - B)

cot} C;

sin } (a+b) these determine 3 (A + B) and 1 (4 B), and thence A and B.

sin a sin C Then c may be found from the formula sin c=

sin A this case, since c is found from its sine, it may be uncertain which of two values is to be given to it; the point may be sometimes settled by observing that the greater side of a triangle is opposite to the greater angle. Or we may determine c from equation (1) of Art. 54, which is free from ambiguity.

Or we may determine c, without previously determining A and B, from the formula cos c = cos a cos b + sin a sin b cos C ; this is free from ambiguity. This formula may be adapted to logarithms thus :

cos c= cos b (cos a + sin a tan b cos C);

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