the arc whose cosine

66

is 9.602000 is 66° 25′ 31′′; cotangent is 10.300000 is 26 37 10.

44. For arcs not exceeding half a degree, it will be most convenient to reverse the method of Art. 40. For this purpose subtract the log. sine of 1" from the given log. sine, and the remainder will be the logarithm of the number of seconds in the arc.

Required the arc whose log. sine is
Subtracting the log. sine of 1"

we have

which is the log. of 206.26.

Hence the required arc is 3′ 26′′.26.

7.000000

4.685575

2.314425,

Required the arc whose log. tangent is 7.500000

Subtracting the log. tangent of 1"

we have

which is the log. of 652.27.

Hence the required arc is 10' 52".27.

4.685575

2.814425,

SOLUTION OF RIGHT-ANGLED TRIANGLES.

THEOREM I.

45. In any right-angled triangle, radius is to the hypothenuse as

the sine of either acute angle is to the opposite side, or the cosine

of either acute angle to the adjacent side.

Let the triangle CAB be right-angled at

A; then will

R: CB :: sin. C : BA : : cos. C : CA. From the point C as a centre, with a radius equal to the radius of the tables, describe the arc DE, and on AC let fall the

perpendicular EF. Then EF will be the sine, and CF the cosine of the angle C.

Because the triangles CAB, CFE are similar, we have

CE: CB:: EF: BA,

or

Also,

or

R: CB:: sin. C: BA.

CE: CB:: CF: CA,

R: CB:: cos. C: CA.

Cor. If radius be taken as unity, we shall have

AB CB sin. C, and AC=CB cos. C.

Hence, in any right-angled triangle, either of the sides which contain the right angle is equal to the product of the hypothenuse by the sine of the angle opposite to that side, or by the cosine of the acute angle adjacent to that side.

THEOREM II.

46. In any right-angled triangle, radius is to either side as the tangent of the adjacent acute angle is to the opposite side, or the secant of the same angle to the hypothenuse.

Let the triangle CAB be right-angled at A; then will

R: CA:: tang. C: AB:: sec. C: CB. From the point C as a centre, with a ra dius equal to the radius of the tables, describe the arc DE, and from the point D Then DF will be the tangent,

draw DF perpendicular to CA. and CF the secant of the angle C.

Because the triangles CAB, CDF are similar, we have

or

Also,

ΟΙ

CD: CA:: DF: AB,

R: CA:: tang. C: AB.
CD: CA:: CF: CB,
R: CA:: sec. C: CB.

Cor. If radius be taken as unity, we shall have
AB AC tang. C, and BC=AC sec. C.

Hence, in any right-angled triangle, either of the sides which contain the right angle is equal to the product of the other side by the tangent of the angle which is opposite to the first side; and the hypothenuse is equal to the product of either side by the secant of the acute angle adjacent to that side.

47. In every plane triangle there are six parts: three sides and three angles. Of these, any three being given, provided one of them is a side, the others may be determined. In a right-angled triangle, one of the six parts, viz., the right angle, is always given; and if one of the acute angles is given, the other is, of course, known. Hence the number of parts to be considered in a right-angled triangle is reduced to four, any two of which being given, the others may be found.

It is desirable to have appropriate names by which to desig nate each of the parts of a triangle. One of the sides adjacent to the right angle being called the base, the other side adjacent to the right angle may be called the perpendicular. The three sides will then be called the hypothenuse, base, and perpendicular. The base and perpendicular are sometimes called the legs of the triangle. Of the two acute angles, that which is adjacent to the base may be called the angle at the base, and the other the angle at the perpendicular.

We may, therefore, have four cases, according as there are given,

1. The hypothenuse and the angles;

2. The hypothenuse and a leg;

3. One leg and the angles; or, 4. The two legs.

All these cases may be solved by the two preceding theorems.

CASE I.

48. Given the hypothenuse and the angles, to find the base and perpendicular.

This case is solved by Theorem I.

Radius: hypothenuse:: sine of the angle at the base: perpendio ular;

Radius: hypothenuse:: cosine of the angle at the base: base. Ex. 1. Given the hypothenuse 275, and the angle at the base 57° 20', to find the base and perpendicular.

The natural sine of 57° 20' is .8418.

1:275::.5398:148.4-AC.

The computation is here made by natural numbers. If we work the proportion by logarithms, we shall have

• radius,

is to the hypothenuse 275,

10.000000

2.439333

Ex. 2. Given the hypothenuse 67.43, and the angle at the perpendicular 38° 43', to find the base and perpendicular.

Ans. The base is 42.175, and perpendicular 52.612. The student should work the examples both by natural num. bers and by logarithms until he has made himself perfectly familiar with both methods. He may then employ either method, as may appear to him most expeditious.

N

CASE II.

49. Given the hypothenuse and one leg, to find the angles and the other leg.

This case is solved by Theorem I.

Hypothenuse: radius::base: cosine of the angle at the base. Radius: hypothenuse:: sine of the angle at the base: perpendicular.

When the perpendicular is given, perpendicular must be substituted for base in this proportion.

Ex. 1. Given the hypothenuse 54.32, and the base 32.11, to find the angles and the perpendicular.

That is, the angle C-53° 45' 47", and therefore the angle B=36°

is to 43.813, the perpendicular 1.641607.

Ex. 2. Given the hypothenuse 332.49, and the perpendicular 98.399, to find the angles and the base.

Ans. The angles are 17° 12' 51" and 72° 47' 9"; the base, 317.6.

CASE III.

50. Given one leg and the angles, to find the other leg and hypothenuse.

This case may be solved by Theorem II.

Radius: base:: tangent of the angle at the base: the perpendicular. :: secant of the angle at the base: hypothenuse.

When the perpendicular is given, perpendicular must be sub stituted for base in this proportion.

This case may also be solved by Theorem I.

sin. B: base:: sin. C; perpendicular;

:: radius: hypothenuse..

Ex. 1. Given the base 222, and the angle at the base 25° 15', to find the perpendicular and hypothenuse.

By natural numbers we have

is to 245.45, the hypothenuse, 2.389966.

Ex. 2. Given the perpendicular 125, and the angle at the perpendicular 51° 19′, to find the hypothenuse and base.

Ans. Hypothenuse, 199.99; base, 156.12.

CASE IV.

51. Given the two legs, to find the angles and hypothenuse. This case is solved by Theorem II.

Base: radius:: perpendicular: tangent of the angle at the base. Radius: base:: secant of the angle at the base: hypothenuse. When the angles have been found, the hypothenuse may be found by Theorem I.

sin. C: AB:: radius: BC.

Ex. 1. Given the base 123, and perpendicular 765, to find the angles and hypothenuse.