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Thus in the right angled triangle ABC, A being the right angle, AC, AB, 90°-B, 90° — BC, 90°-C, are the circular parts, by Def. 1.; and if any one as AC be reckoned the middle part, then AB and 900 -C, which are contiguous to it on different sides, are called adjacent parts ; and 90°-B, 90° – BC are the opposite parts. ln like manner

B В

A

if AB is taken for the middle part, AC and 90° – B are the adjacent parts : 90°. BC, and 90°-C are the opposite. Or if 90° BC be the middle part, 90-B, 90°- Care adjacent ; AC and AB opposite, &c.

This arrangement being made, the rule of the circular part is contained in the following

PROPOSITION.

In a right angled spherical triangle, the rectangle under the radius and

the sine of the middle part, is equal to the rectangle under the tangents of the adjacent parts; or to the rectangle under the cosines of the opposite parts.

The truth of the two theorems included in this enunciation may be easily proved, by taking each of the five circular parts in succession for the middle part, when the general proposition will be found to coincide with some one of the analogies in the table already given for the resolution of the cases of right angled spherical triangles. Thus, in the triangle ABC, if the complement of the hypotenuse BC be taken as the middle part, 90° B, and 90°-C, are the adjacent parts, AB and AC the opposite. Then the general rule gives these two theorems, R Xcos BC=cot B Xcot C, and R Xcos BC=cos AB Xcos AC. The former of these coincides with the cor. to the 20th ; and the latter with the 22d.

To apply the foregoing general proposition, to resolve any case of a right angled spherical triangle, consider which of the three quantities named (the two things given and the one required) must be made the middle term, in order that the other two may be equidistant from it, that is, may be both adjacent, or both opposite ; then one or other of the two theorems contained in the above enunciation will give the value of the thing required.

Suppose, for example, that AB and BC are given, to find C; it is evident that if AB be made the middle part, BC and C are the oppos

site parts, and therefore R Xsin AB=sin C X sin BC, for sin C=cos (90° --C), and cos (90°-BC)=siu BC, and consequently

sin C

sin AB sin BC

cos C

Again, suppose that BC and C are given to find AC; it is obvious that C is in the middle between the adjacent parts AC and (90° - BC), therefore RXcos C=tan AC Xcot BC, or tan AC=.

:coș C+ cot BC

1 tan BC; because, as has been shewn above, =tan BC,

'cot BC In the same way may all the other cases be resolved. One or two trials will always lead to the knowledge of the part which in any given case is to be assumed as the middle part; and a little practice will make it easy, even without such trials, to judge at once which of them is to be so assumed. It may be useful for the learner to range the names of the five circular parts of the triangle round the circumference of a circle, at equal distances from one another, by which means the middle part will be immediately determined.

Besides the rule of the circular parts, Napier derived from the last of the three theorems ascribed to him above, (schol. 29.), the solutions of all the cases of oblique angled triangles. These solutions are as follows: A, B, C, denoting the three angles of a spherical triangle, and a, b, c, the sides opposite to them.

I.
Given two sides b, c, and the angle A between them.

To find the angles B and C.

sin (b-) tan (B-C)=cot AX

(31.) cor. 1. sinį (b+c)

cos } (b-c) tan (B+C)=cot AX

(31.) cor. 1.
cos (b + c)
To find the third side a.
sin B : sin A :: sin b : sin a.

II. Given the two sides b, c, and the angle B opposite to one of them.

To find C, and the angle opposite to the other side.

sin b : sin c :: sin B : sin C.

To find the contained angle A.

sin 4 (b+c) Cot 4 A=tan } (B-C) X

sin ; (6-4)

(31.) cor. 1.

To food the third side a.

Sin B :: sin A:: sin b : sin a.

III.
Given two angles A and B, and the side c between them.

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Given two angles A and B, and the side a, opposite to one of them.

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The other two cases, when the three sides are given to find the an. gles, or when the three angles are given to find the sides, are resolved by the 29th, (the first of NAPIER’s Propositions,) in the same way as in the table already given for the cases of the oblique angled triangle.

There is a solution of the case of the three sides being given, which it is often very convenient to use, and which is set down here, though the proposition on which it depends bas not been demonstrated,

Let a, b, o be the three given sides, to find the angle A, contained between b and c.

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In like manner, if the three angles, A, B, C are given to find c, the şide between A and B.

Let A + B + C=S,

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These theorems, on account of the facility with which Logarithms are applied to them, are the most convenient of any for resolving the two cases to which they refer. When A is a very obtuse 'angle, the second theorem, which gives the value of the coside of its is to be used; otherwise the first theorem, giving the value of the sine of its half is preferable. The same is to be observed with respect to the side 6, the reason of which was explained, Plane Trig. Schel.

END OF SPHERICAL TRIGONOMETRY.

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