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It remains to shew 'what will be the result on the supposition that one side of the right-angle is greater, and the other less than 90°.

For example, let AB be greater than the quadrant AD; and AC less than AD: we shall have the arc CD, 90°, and the angle CDA will be acute, therefore CDB will be obtuse, and consequently the arc BC, opposite to it, will be greater than

B CD, and therefore more than 90°.

77. From what we have shewn, it follows that in a spherical triangle, if the two sides of the right-angle are of the same species, the hypoihenuse will be less than 90°; and if they are of different species, the hypothenuse will be greater than 90o.

And as the oblique angles are always of the same species as the sides opposite to them, they equally serve to shew of what species the hypothenuse is.

78. And reciprocally the hypothenuse may serve to shew of what species are the sides of the right and oblique angles. If, for example, the hypothenuse, and one of the sides be of the same species, the other side is less than 90°; and if the hypothenuse and one of the sides be of different species, the other side is more than 90°.

Observe, however, that in the case where one of the sides of the right-angle is 90°, it may happen likewise that the other side is 90°, as well as the first. In this case the three angles are evidently right-angles. It may also happen that this other side may be more or less than 90°. We shall explain the mode of solving spherical triangles in these different cases.

79. In plane trigonometry it has been shewn that the sines of the angles are proportional to the sides opposite these same angles. But as in spherical triangles, the sides are circular arcs, the same proportion cannot obtain here. We shall endeavour to substitute another which applies to all sorts of triangles.

PRINCIPLES AND PROPORTIONS FOR THE SOLUTION OF SPHERICAL TRIANGLES.

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Drawing the line DI, it is evident that the triangle CDI, rightangled at I, is similar to the triangle FGE, right-angled in G; we have therefore

FE: CD::FG: CI that is, radius is to the sine of the hypothenuse, as the sine of the angle B is to the sine of the opposite arc CA. In like manner it may be proved that radius is to the sine of the hypothenuse, as the sine of the angle C is to the sine of the opposite arc AB. Generally, therefore,

81. In every right-angled spherical triangle, radius is to the sine of the hypothenuse, as the sine of one of the oblique angles is to the sine of the side opposite to it.

82. If the trian. gle ABC is obliqueangled, letting fall the perpendicular are CD, we shall have the two following

pro-
A

A А

B portions; by (81)

B D
R: sin AC::sin A : sin CD

R: sin BC::sin B : sin CD
Therefore,

sin A : sin B::sin BC : sin AC and consequently in any spherical triangle, the sines of the angles are proportional to the sines of the sides.

83. Suppose now the triangle ,ABC to be right-angled at A, and the sides BC and AC

E to be prolonged to 90°, the one in D and the other in F; if we draw the arc DE, we shall have the triangle CDE, right-angled in E, whose four parts CE, CD, DE, and D, are respectively the complements of the four BC, AC, B, and AB, of the triangle ABC, as may easily be shewn. Triangles formed in this manner are called complemental triangles. Now

B in the complemental triangle CDE, we have

R: sin CD::sin D : sin CE

R: cos AC::cos AB : cos BC Therefore in any right-angled spherical triangle, radius is to the cosine of one of the sides of the right-angle, as the cosine of the other side is to the cosine of the hypothenuse.

And consequently, if an oblique-angled spherical triangle be divided into two right-angled triangles. by an arc perpendicular to its base, we shall always have the cosines of the segments of the base proportional to the cosines of the two adjacent sides.

So that in the triangle ABC, for example, after having described

or,

:: cota

the perpendicular arc CD, we shall have cos AC : cos BC::cos AD: cos BD.

This proportion produces this subsequent one, cos AC +cos BC : cos BCcos AC::cos AD+cos BD : cos BDcos AD, and from this we easily deduce

B this third proportion (19 Pl. Trig.)

AC + BC BC-AC AD+BD BD-AD
cot
: tang -

: tang-
2
2
2

2 But AD+DB-AB, when the perpendicular falls within the triangle; by substituting tangents for cotangents, we have therefore the following theorem of such frequent use, and which may be announced as follows:

84. In any oblique-angled spherical triangle, on the base of which we have let fall a perpendicular arc, and which lies within the trian, gle, the tangent of the half base is to the tangent of half the sum of the other two sides, as the tangent of the half difference of these sides is to the tangent of half the difference of the segments of the base.

Observe that if the arc falls without the triangle, we should have AD-BA -AB; and we must then merely substitute the tangent of half the sum of the segments, to the tangent of the half AS difference of these segments.

B 85. Returning to the complemental triangle CDE, we have

R: sin CD::sin C: sin DE E Therefore R: cos AC::sin C : cos B

That is to say, in any right-angled spherical triangle, radius is to the cosine of one of the sides of the right-angle, as the sine of the oblique angle opposite to the other side is to the cosine of the other oblique angle. 86. And consequently if we let fall an arc A

B perpendicularly on the base of an oblique-angled triangle, the sines of the angles at the summit will be proportional to the cosines of the angles of the base. In the oblique-an

с gled triangle ACB, we have therefore sin ACD : sin BCD:: cos A : cos B

B

B D

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G H

87. Let there now be the triangle ABC, right-angled at A, and let there be drawn the tangents BP

Q and BQ: the latter to the hypothenuse BC, and the first to the side BA. If we draw the secants EP and EQ,

F whose extremities are joined by the line PQ, we shall have the triangles PEQ and BPQ, włose planes are perpendi- E

B cular to that of the base BAE. Their

А intersection BQ will consequently be perpendicular to the same plane; and the triangle_BPQ, rightangled at P, will be similar to the triangle CFE. Thus we shall have

FE: GE :: BQ : BP or,

Ꭱ ;

: cos B :: tang BC : tang AC Therefore in any right-angled spherical triangle, radius is to the cosine of one of the oblique angles, as the tangent of the hypothenuse

to the tangent of the side opposite to the other angle.

88. If the triangle ABC is oblique

C angled, we may let fall on its base the perpendicular arc CD, which will A

A give

B
D

B

R: cos ACD :: tang AC : tang CD and

R: cos BCD ; : tang BC : tang CD Whence we infer,

cos ACD : cos BCD :: tang BC : tang AC That is, if we let fall an arc perpendicularly on the base of an oblique-angled spherical triangle, the cosines of the angles at the summit will always be reciprocally proportional to the tangents of the adjacent sides.

89. And as in the complemental triangle CDE, we have R: cos D:: tang CD : tang DE

D It is evident that we also have

E R: sin AB :: cot AC :cot B::tang B: tang AC

Therefore, in any right-angled triangle, radius is to the sine of one of the sides of the right-angle, as the tangent of the oblique angle opposite to the other side is to the tangent of this last side. 90. The same triangle CDE gives R:sin CE:: tang C : tang DE

B therefore, R: cos BC ;: tang C: cot B

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