PROB. V.

To find the logarithms of √2, √3, √5, &c., √20, √30, &c., †2, &c. Since √2x2=2; log√2+log√2=log 2.

Now,

log 2=.301...log√2=.1505

Similarly, log√3=.2385; log√5=.34947 Again, log 20-1.301.. log/20=.6505 log 30=1.477.. log/30=.7385

Lastly,

2×2×2=2.. 3 log2=log 2=.301

and log 2.10034.

PROB. VI.

To explain the logarithms of trigonometrical functions. These are better found by algebraic formulæ, but an idea of their values and nature may be obtained by the following examples.

It must be premised, that the radius adopted in logarithmic tables is 1010, so that its logarithm is 10. Hence the logarithms of the sines of tolerably small angles are positive, which (Cor. 3, Prop. 1) would not be the case were the radius unity.

Thus the logarithms of sin 45° is (Art. 15), log

1010

√2

1010

=

2

· 10 – log 2 (Art. 16) = 9.699

The trigonometric functions of all angles from 1' to 90° are given in tables to every minute, and these, with the logarithms of numbers, constitute the most important portion of a volume of logarithmic tables. With such tables, the solution of triangles is a matter of simple addition and subtraction.

PROB. VII.

To solve triangles by the aid of logarithms.

1st, Without the tables, simply as an illustration.

1. Two angles of a triangle are 30° and 45°, and the side opposite the latter is 10/2, what is the side opposite to the former? we have sin 45°: sin 30°:: 10/2 side required

.. log side required = log 102 + log sin 30°- log sin 45°

and the side required = 10.

2. Find the area of the triangle of which the sides are 13, 12, and 5 feet.

The result of Art. 49 is usually expressed in the following words:

Add the three sides together, and take half their sum; from the half sum subtract each side separately; multiply the half sum and the three remainders continually together, and extract the square root of the product for the area.

Here the half sum is 15, and the remainders are 2, 3, and 10; hence the area = √15×2×3×10

1

and log area = {log 15+ log 2 + log 3+ log 10} (Def.)

=

=

2

1

{log 3+ log 5+ log 2 + log 3 + log 10}

·{ . 477 + . 699 + . 301 +. 477 +1} (Prob. 5.)

2d, By the tables.

3. The two sides of a right-angled triangle are √20 and √30, what are the angles?

tan C: R::AB: AC (33)

4. A person on the top of a hill observes two milestones on the plain, lying in a right line from the hill; the angle of depression of the nearer is 40°, and of the farther 25°; required the height

of the hill.

We have, distance of nearer milestone: 1 mile :: sin 15° : sin 25°; and height of hill: distance of nearer milestone:: sin 40° : R. sin 15° sin 40° sin 25°

hence, height 1760

=

R

in yards

log height = log 1760+log sin 15°+log sin 40°

-log 25°-10

ELEMENTS OF SPHERICAL TRIGONOMETRY.

SECTION I.

PROP. I.

If a sphere be cut by a plane through the centre, the section is a circle, having the same centre with the sphere, and equal to the circle by the revolution of which the sphere has been described.

For all the straight lines drawn from the centre to the superficies of the sphere are equal to the radius of the generating semicircle (III. Sup. Def. 7); hence the common section of the spherical superficies, and of a plane passing through the centre, is a line lying in one plane, and having all its points equally distant from the centre of the sphere; therefore it is the circumference of a circle (I. Def. 11), having for its centre the centre of the sphere, and for its radius the radius of the sphere; that is, of the semicircle by which the sphere has been described. It is equal, therefore, to the circle of which that semicircle was a part. `Q. E. D.

DEFINITIONS.

I. Any circle which is a section of a sphere by a plane passing through its centre, is called a great circle of the sphere.*

COR. 1. All great circles of a sphere are equal (Prop. 1.) COR. 2. Any two great circles bisect one another. For the straight line (II. Sup. 3) in which their planes intersect one another, passes through both their centres.

II. The pole of a great circle is that point in which a perpendicular to its plane from the centre of the sphere meets the sphere.

COR. Every great circle has two poles at the opposite extremities of a diameter.

III. A spherical angle is the angle contained by the arcs of two great circles, and is the same with the angle contained by the tangents to the arcs at their point of intersection.

We shall frequently call it simply a circle, inasmuch as none but great circles occur in this science.

COR. It is equal to the angle contained by their planes (II. Sup.
Def. 4).
IV. The side of a spherical triangle is that arc of a great circle
intercepted between two others, which is less than a semicircle.

PROP. II.

If a point on the surface of a sphere be distant by a quadrant from each of two points in a given great circle, not at opposite extremities of a diameter, it is the pole of that circle.

Let the point A be distant by a quadrant from each of the points B, C, in the circle BC, not at the opposite extremities of a diameter; A is the pole of BC.

Let D be the centre of the sphere; join DA, DB, DC.

Then, because AB, AC are quadrants, the angles ADB, ADC are right angles; therefore DA is at right angles to the plane BDC (II. Sup. 4); and A is the pole of BC (Def. 2). Q. E. D.

PROP. III.

B

If the pole of a great circle be the same with the intersection of other two great circles, the arc of the first-mentioned circle intercepted between the other two is the measure of the spherical angle which the same two circles make with one another.

Let the great circles BA, CA on the superficies of a sphere, of which the centre is D, intersect one another in A, and let BC be an arc of another great circle, of which the pole is A; BC is the measure of the spherical angle BAC.

E

D

A

Join DA, DB, DC, and draw AE, AF tangents at A. Then because A is the pole of BC, the angles_ADB, ADC are right angles; and DAE, DAF are also right angles (III. 15); therefore (I. 29) AE is parallel to BD, and AF to DC; consequently (II. Sup. 9) the angle BDC is equal to the angle EAF, that is, to the spherical angle BAC (Def. 3). Therefore the B arc BC is the measure of the spherical angle BAC. Q. E. D.

PROP. IV.

If the planes of two great circles of a sphere be at right angles to one another, the circumference of each of the circles passes through the poles of the other; and if the circumference of one great circle pass through the poles of another, the planes of these circles are at right angles.

Let ACBD, AEBF be two great circles, the planes of which are at right angles to one another; the poles of the circle AEBF

are in the circumference ACBD, and the poles of the circle ACBD in the circumference AEBF.

F

C

From G the centre of the sphere, draw GC, GE in the planes ACBD, AEBF respectively, perpendicular to AB. Then CGE is the angle contained by the planes (II. Sup. Def. 4), and is therefore a right angle. Consequently, CG is at right angles to both GB and GE, and therefore (II. Sup. 4) CG is at right angles to the plane AEBF; hence A (Def. 2) C is the pole of the circle AEBF; and if CG be produced to D, D is the other pole of the circle EB.

D

B

E

In the same manner it is shown that E and F are the poles of the circle CB. Therefore, the poles of each of these circles are in the circumference of the other.

Again, if C be one of the poles of the circle EB, the great circle CB which passes through C is at right angles to the circle EB. For CG being drawn from the pole to the centre of the circle EB is at right angles to the plane of that circle (Def. 2); and, therefore, every plane passing through CG (II. Sup. 17) is at right angles to the plane AEBF; now, the plane ACBD passes through CG. Therefore, &c. Q. E. D.

COR 1. If of two great circles the first passes through the poles of the second, the second also passes through the poles of the first. For, if the first passes through the poles of the second, the plane of the first must be at right angles to the plane of the second, by the second part of this proposition; and, therefore, by the first part of it, the circumference of each passes through the poles of

the other.

COR. 2. All great circles that have a common diameter have their poles in the circumference of a circle, the plane of which is perpendicular to that diameter.

PROP. V.

The angle subtended at the centre of the sphere by the poles of two great circles is equal to the angle between the circles themselves.

F

E

Let AB, AC be two great circles, EF their poles, D the centre of the sphere; then, if DE, DF be joined, the angle EDF is equal to the spherical angle A. Let the plane FDE cut the circles AB, AC in B and C, join DB, DC. Because E is the pole of AB, the angles EDA, EDB are right angles; and because F is the pole of AC, the angles FDA, FDC are right angles. Hence, both the angles EDA, FDA are right angles, and therefore (II. Sup. 4) DA is

at right angles to the plane FDE. Hence, ADC, ADB are right