Making the base radius.

R:AB:: Sec A: AC
R:AB:: Tan A: BC

Making the perpendicular radius.

Tan C:AB::R:BC

R:BC:: Sec C: AC

138. Ex. 2. If the perpendicular (Fig. 21.) be 74, and the angle C 61° 27', what is the length of the base and the hypothenuse?

Making the hypothenuse radius.

Sin A: BC::R: AC
R:AC:: Sin C: AB

Ex. 1. If the base (Fig. 22,) be 284, and the perpendicular 192, what are the angles, and the hypothenuse?

In this case, one of the legs must be made radius, to find an angle; because the hypothenuse is not given.

Ex. 2. If the base be 640, and the perpendicular 480, what are the angles and hypothenuse?

Ans. The hypothenuse is 800, and the angle at the base 36° 52′ 12′′.

Examples for practice.

1. Given the hypothenuse, 68, and the angle at the base 39° 17'; to find the base and perpendicular.

2. Given the hypothenuse 850, and the base 594, to find the angles, and the perpendicular.

3. Given the hypothenuse 78 and perpendicular 57, to find the base, and the angles.

4. Given the base 723, and the angle at the base 64° 18', to find the hypothenuse and perpendicular.

5. Given the perpendicular 632, and the angle at the base 81° 36', to find the hypothenuse and the base.

6. Given the base 32, and the perpendicular 24, to find the hypothenuse, and the angles.*

140. The preceding solutions are all effected, by means of the tabular sines, tangents, and secants. But, when any two sides of a right angled triangle are given. the third side may be found, without the aid of the trigonometrical tables, by the proposition, that the square of the hypothenuse is equal to the sum of the squares of the two perpendicular sides. (Euc. 47. 1.)

If the legs be given, extracting the square root of the sum of their squares, will give the hypothenuse. Or, if the hy

* See note F.

pothenuse and one leg be given, extracting the square root of the difference of the squares, will give the other leg.

Let h=the hypothenuse

p=the perpendicular of a right angled triangle.
b=the base

Ex. 1. If the base is 32, and the perpendicular 24, what is the hypothenuse?

2. If the hypothenuse is 100, and the base 80, what is the perpendicular?

The square of 100 is 10000
of 80

6400

The difference of the squares is 3600
The root of which is

60, the perpendicular.

3. If the hypothenuse is 300, and the perpendicular 220, what is the base?

Ans. 300-220* =41600, the root of which is 204 nearly.. —¡→→

141. It is generally most convenient to find the difference of the squares by logarithms. But this is not to be done by subtraction. For subtraction, in logarithis, performs the of fice of division. (Art. 41.) If we subtract the logarithm of b from the logarithm of h2, we shall have the logarithm, not of the difference of the squares, but of their quotient. There is, however, an indirect, though very simple method, by which the difference of the squares may be obtained by logarithms. It depends on the principle, that the difference of the squares of two quantities is equal to

J

2

the product of the sum and difference of the quantities. (Alg. 235.) Thus

h2 —b2=(h+b)× (h—b)

as will be seen at once, by performing the multiplication. The two factors may be multiplied by adding their logarithms. Hence,

142. To obtain the difference of the squares of two quantities, add the logarithm of the sum of the quantities, to the logarithm of their difference. After the logarithm of the difference of the squares is found; the square root of this difference is obtained, by dividing the logarithm by 2. (Art. 47.)

Ex. 1. If the hypothenuse be 75 inches, and the base 45, what is the length of the perpendicular?

2. If the hypothenuse is 135, and the perpendicular 108,

what is the length of the base?

Ans. 81.

SECTION 1V.

SOLUTIONS OF OBLIQUE ANGLED TRIANGLES.

ART. 143. THE sides and angles of oblique angled triangles may be calculated by the following the

orems.

THEOREM I.

In any plane triangle, the sines of the angles are as their opposite sides.

Let the angles be denoted by the letters A, B, C, and their opposite sides by a, b, c, as in Fig. 23 and 24. From one of the angles, let the line p be drawn perpendicular to the opposite side. This will fall either within or without the triangle.

1. Let it fall within as in Fig. 23. Then, in the right angled triangles ACD and BCD, according to art. 126.

R:b::Sin A:p
R:a:: Sin B:p.

Here, the two extremes are the same in both proportions. The other four terms are, therefore, reciprocally proportional: (Alg. 387.*) that is,

a:b::Sin A: Sin B.

2. Let the perpendicular p fall without the triangle, as in Fig. 24. Then, in the right angled triangles ACD and BCD;

R:b:: Sin A:
:P

R:a: Sin B: :P

Therefore as before,

a:b::Sin A: Sin B.

Sin A is here put both for the sine of DAC, and for that

*Euclid 23. 5.