Ex. 2. Given the base 53, and perpendicular 67, to find the angles and hypothenuse.

Ans. The angles are 51° 39′ 16′′, and 38° 20′ 44′′; hypothenuse, 85.428..

Examples for Practice.

1. Given the base 777, and perpendicular 345, to find the hypothenuse and angles.

This example, it will be seen, falls under Case IV.

2. Given the hypothenuse 324, and the angle at the base 48° 17', to find the base and perpendicular.

3. Given the perpendicular 543, and the angle at the base 72° 45', to find the hypothenuse and base.

4. Given the hypothenuse 666, and base 432, to find the angles and perpendicular.

5. Given the base 634, and the angle at the base 53° 27', to find the hypothenuse and perpendicular.

6. Given the hypothenuse 1234, and perpendicular 555, to find the base and angles.

7. Suppose the radius of the earth to be 3963 miles, and that it subtends an angle of 57' 2".3 at the moon, what is the distance of the moon from the earth?

8. Suppose that when the moon's distance from the earth is 238,885 miles, its apparent semi-diameter is 15' 33".5, what is its diameter in miles?

9. Suppose the radius of the earth to be 3963 miles, and that it subtends an angle of 8".9 at the sun, what is the distance of the sun from the earth?

10. Suppose that the sun's distance from the earth is 92,000,000 miles, and that its apparent semi-diameter is 16' 1".8, what is its diameter in miles?

52. When two sides of a right-angled triangle are given, the third may be found by means of the property that the square of the hypothenuse is equal to the sum of the squares of the other two sides.

Hence, representing the hypothenuse, base, and perpendicular by the initial letters of these words, we have

h=√/b2+p2; b=√√/h2—p2; p=√.h2—b2.

Ex. 1. If the base is 2720, and the perpendicular 3104, what is the hypothenuse?

Ans. 4127,1.

Ex. 2. If the hypothenuse is 514, and the perpendicular 432, what is the base?

SOLUTION OF OBLIQUE-ANGLED TRIANGLES.

THEOREM I.

53. In any plane triangle, the sines of the angles are proportional to the opposite sides.

Let ABC be any triangle, and from one of its angles, as A, let AD be drawn perpendicular to the opposite side BC. There may be two cases.

D

First. If the perpendicular falls within the trian-B
gle, because the triangle ABD is right-angled at D, we have
R: sin. B:: AB: AD; whence Rx AD-sin. Bx AB.
For a similar reason,

R: sin. C :: AC: AD; whence RxAD=sin. C × AC.
sin. Bx AB=sin. Cx AC;
sin. B: sin. C:: AC: AB.

Therefore

or,

Second. If the perpendicular falls without the triangle, we have in the triangle ABD, as before,

But, since ABD is the supplement of ABC, their sines are equal,

Art. 13.

Therefore

sin. ABC: sin. C:: AC: AB.

THEOREM II.

54. In any plane triangle, the sum of any two sides is to their difference as the tangent of half the sum of the opposite angles is to the tangent of half their difference. Let ABC be any triangle; then will

CB+CA: CB-CA:: tang.

A+B
2

A-B

:tang.

2

Produce AC to D, making CD equal to CB, and join DB. Take CE equal to CA; draw AE, and produce it to F. Then AD is the sum of CB and CA, and BE is their difference.

The sum of the two angles CAE, CEA is equal to the sum of CAB, CBA, each being the supplement of ACB (Geom., B. I., Pr. 27). But, since CA is equal to CE, the angle CAE is equal to the angle CEA; therefore CAE is the half sum of the angles CAB,

E

A

B

F

CBA. Also, if from the greater of the two angles CAB, CBA there be taken their half sum, the remainder, FAB, will be their half difference (Algebra, p. 89).

Since CD is equal to CB, the angle ADF is equal to the angle EBF; also, the angle CAE is equal to AEC, which is equal to the vertical angle BEF. Therefore the two triangles DAF, BEF are mutually equiangular; hence the two angles at F are equal, and AF is perpendicular to DB.

If, then, AF be made radius, DF will be the tangent of DAF, and BF will be the tangent of BAF. But, by similar triangles, we have

55. If from any angle of a triangle a perpendicular be drawn to the opposite side or base, the sum of the segments of the base is to the sum of the two other sides as the difference of those sides is to the difference of the segments of the base.

For demonstration, see Geometry, B. IV., Pr. 34, Cor.

56. In every plane triangle three parts must be given to enable us to determine the others, and of the given parts one at least must be a side. For, if the angles only are given, these might belong to an infinite number of different triangles. In solving oblique-angled triangles four different cases may therefore be presented. There may be given,

1. Two angles and a side;

2. Two sides and an angle opposite one of them;

3. Two sides and the included angle; or,

4. The three sides.

We shall represent the three angles of the proposed triangle by A, B, C, and the sides opposite them respectively by a, b, c.

CASE I.

57. Given two angles and a side, to find the third angle and the other two sides.

To find the third angle, add the given angles together, and subtract their sum from 180°.

The required sides may be found by Theorem I. The propor tion will be,

The sine of the angle opposite the given side: the given side ::the sine of the angle opposite the required side

:the required side.

Ex. 1. In the triangle ABC, there are given the angle A, 57° 15', the angle B, 35° 30', and the side c, 364, to find the other parts.

The sum of the given angles, subtracted from 180°, leaves 87° 15' for the angle C. Then, to find the side a, we say,

By natural numbers,

sin. Cc:: sin. A: a.

A

.9988:364::.8410:306.49=a.

This proportion is most easily worked by logarithms, thus: As the sine of the angle C, 87° 15', comp. 0.000500

To find the side b, we have, sin. C: c:: sin. B: b.

By natural numbers,

9988:364.5807:211.62=b.

The work by logarithms is as follows:

2.561101

9.924816

2.486417.

B

Ex. 2. In the triangle ABC, there are given the angle A, 49° 25', the angle C, 63° 48', and the side c, 275, to find the other parts. Ans. B 66° 47'; a=232.766; b=281.67.

CASE II.

58. Given two sides and an angle opposite one of them, to find the third side and the remaining angles.

One of the required angles is found by Theorem I. The proportion is,

The side opposite the given angle: the sine of that angle :: the other given side: the sine of the opposite angle.

The third angle is found by subtracting the sum of the other two from 180°; and the third side is found as in Case I.

A B

C

Β'

If the side BC, opposite the given angle A, is shorter than the other given side AC, the solution will be ambiguous; that is, two different triangles ABC, AB'C may be formed, each of which will satisfy the conditions of the problem.

The numerical result is also ambiguous, for the fourth term of the first proportion is a sine of an angle. But this may be the sine either of the acute angle AB'C, or of its supplement, the obtuse angle ABC (Art. 13). In practice, however, there will generally be some circumstance to determine whether the required angle is acute or obtuse. If the side opposite the given angle is longer than the other given side, there can be no ambiguity, for B will fall on B'A produced, and the triangle ABC will no longer be one solution of the problem. This is always the case when the given angle is obtuse.

Ex. 1. In a triangle ABC, there are given AC, 458, BC, 307, and the angle A, 28° 45', to find the other parts.

To find the angle B;

BC: sin. A:: AC: sin. B.

By natural numbers,

307.4810:458:.7176, sin. B, the arc corresponding to which is 45° 51', or 134° 9'.

This proportion is most easily worked by logarithms, thus:

The angle ABC is 134° 8′ 46," and the angle AB'C, 45° 51′ 14′′. Hence the angle ACB is 17° 6' 14", and the angle ACB', 105° 23' 46".