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

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

Ans. 40.

Ans. 60.

2. If the hypothenuse is 100, and the base 80, what is the perpendicular? 3. If the hypothenuse is 300, and the perpendicular 220, what is the base ?

Ans. 300-220-4160, 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 logarithms, performs the office of division. (Art. 41.) If we subtract the logarithm of b2 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 the product of the sum and difference of the quantities. (Alg. 191.) 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?

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2. If, the hypothenuse is 135, and the perpendicular 108, what is the length of the base?

Ans. 81.

SECTION IV.

SOLUTIONS OF OBLIQUE ANGLED TRIANGLES.

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

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

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of the angles, let the line p be drawn perpendicular to the opposite side. This will fall either within or without the triangle.

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

Rb 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: that is,

ab 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;

Rb sin A P

R: a :: sin B: p

Therefore, as before,

ab sin A: sin B.

Sin A is here put both for the sine of DAC, and for that of BAC. For, as one of these angles is the supplement of the other, they have the same sine. (Art. 90.)

The sines which are mentioned here, and which are used

* Euclid, 23. 5.

4

in calculation are tabular sines. But the proportion will be the same, if the sines be adapted to any other radius. (Art. 119.)

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THEOREM II.

144. In a plane triangle,

AS THE SUM OF ANY TWO OF THE SIDES,

TO THEIR DIFFERENCE;

So IS THE TANGENT OF HALF THE SUM OF THE

OPPOSITE ANGLES;

TO THE TANGENT OF HALF THEIR DIFFERENCE.

Thus, the sum

of AB and AC, is to their difference; as the tangent of half the

sum of the an

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gles ACB and

ABC, to the tan

gent of half their difference.

B

Demonstration.

Extend CA to G, making AG equal to AB; then CG is the sum of the two sides AB and AC. On AB, set off AD, equal to AC; then BD is the difference of the sides AB and AC.

The sum of the two angles ACB and ABC, is equal to the sum of ACD and ADC; because each of these sums is the supplement of CAD. (Art. 79.) But as ACAD by construction, the angle ADC=ACD (Euc. 5 1.*) Therefore ACD is half the sum of ACB and ABC. As AB=AG, the angle AGB=ABG, or DBE. Also, GCE, or ACD= ADC=BDE. (Euc. 15. 1.f) Therefore in the triangles

* Thomson's Legendre, 11. 1.

† Ibid. 4. 1.

GCE, and DBE,

the two remain

ing angles DEB,

and CEG, are equal; (Art. 79.) So that CE is perpendicular to

BG. (Euc. Def.

E

B

7. 1.*) If then CE is made radius, GE is the tangent of GCE, (Art. 84.) that is, the tangent of half the sum of the angles opposite to AB and AC.

If from the greater of the two angles ACB and ABC, there be taken ACD their half sum; the remaining angle ECB will be their half difference. The tangent of this angle, CE being radius, is EB, that is, the tangent of half the difference of the angles opposite to AB and AC. We have then, CG the sum of the sides AB and AC;

DB their difference;

GE the tangent of half the sum of the opposite angles; EB the tangent of half their difference.

But by similar triangles,

CG DB:: GE : EB.

Q. E. D.

THEOREM III.

145. If upon the longest side of a triangle, a perpendicular be drawn from the opposite angle;

AS THE LONGEST SIDE,

TO THE SUM OF THE TWO OTHERS;
So IS THE DIFFERENCE OF THE

LATTER,

TO THE DIFFERENCE OF THE SEG-
MENTS MADE BY THE PERPEN-
DICULAR.

• Thomson's Legendre, Def. 12, 1.

26.

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