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47 and minutes corresponding to this difference in the sin. column will be the angle A required. The other parts will be found as in (1).

Given the Hypotenuse (6) and Angle A, to find the Base (c), and

Perpendicular (a) Example.—Let the hypotenuse 370 yards, and the angle A 56° 30'; required the angle C, also the base, and the perpendicular.

BY CONSTRUCTION
To find Angle C

90° 00

LA56 30
L C33 30

с

ь

Draw the line A B of any length, and make the angle at A 56° 30' (Problem XII. Geometry); from A to C lay off 370, the length of the hypotenuse, taken from any convenient scale of equal parts, and from the point Clet fall the perpendicular CB (Problem III. Geometry); then A B C is the triangle required : the base AB, measured on the same scale of equal parts by which the hypotenuse was measured, will be 204:2, and the perpendicular B C 308.5.

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Calculation by Logarithms
To find the Base.

To find the perpendicular.

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Ans. C= 33° 30'; base = 204:22 yards; perpendicular

308.54 yards.

Given the Base (c), and Perpendicular (a), to find the Angles

and the Hypotenuse (6) Example.Let the base = 35.5, and the perpendicular 41•6; required the angles A and C, and the hypotenuse.

BY CONSTRUCTION.

41.6

Draw the line B A, and from B erect the perpendicular B C; make B A equal to 35.5, and B C equal to 41.6, and draw the line AC; then the hypotenuse AC will measure 547, the angle A 491°, and the angle C 401°.

900

35.5

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Examples for Practice 1. Given the hypotenuse 108, and the angle opposite the perpendicular 25° 36' ; required the base and perpendicular.

Ans. The base is 974, and the perpendicular 46.66.

2. Given the base 96, and its opposite angle 71° 45'; required the perpendicular and the hypotenuse.

Ans. The perpendicular is 31.66, and the hypotenuse 10I•I.

3. Given the perpendicular 360, and its opposite angle 58° 20'; required the base and the hypotenuse.

Ans. The base is 222 and the hypotenuse 423.

4. Given the base 720, and the hypotenuse 980 ; required the angles and the perpendicular.

Ans. The angles are 47° 17' and 42° 43', and the perpendicular 664-8.

5. Given the perpendicular 110-3, and the hypotenuse' 176-5; required the angles and the base.

Ans. The angles are 38° 41' and 51° 19', and the base 137.8.

6. Given the base 360, and the perpendicular 480 ; required the angles and the hypotenuse.

Ans. The angles are 53° 8' and 36° 52', and the hypotenuse 600.

7. Given the base 346-5, and the adjacent angle 35° 24' ; required the perpendicular and the hypotenuse.

Ans. The perpendicular is 246-2, and the hypotenuse 425.1.

8. Given the hypotenuse 36-5, and the angle opposite the base 65° 15'; required the perpendicular and the base.

Ans. The perpendicular is 15•28, and the base 33•15.

9. Given the perpendicular 725, and the adjacent angle 21° 36'; required the base and the hypotenuse.

Ans. The base is 287.1, and the hypotenuse 779.8.

10. Given the base 32.76, and the hypotenuse 56.95; required the angles and the perpendicular.

Ans. The angles are 35° 7' and 54° 53', and the perpendicular 46-58.

II. Given the perpendicular 98-4, and the hypotenuse 1013; required the angles and the base.

Ans. The angles are 5° 34' and 84° 26', and the base 1009.

12. Given the base 4567, and the perpendicular 3251 ; required the angles and the hypotenuse. Ans. The angles are 35° 27' and 54° 33', and the hypotenuse 5606.

OBLIQUE-ANGLED TRIGONOMETRY NOTE.—If you have carefully studied the Ratios (p. 45 you can easily understand the following investigation.

It is required to find a relation between the sides of a plane triangle and the trigonometrical ratios of its angles.

In the triangles A B C, which has all its angles acute angles, let the side opposite angle A be a, that opposite angle B be b, and that opposite angle C be c. From C draw C D as a perpendicular, which call p; then in the right-angled triangle ACD we have

P
sin. A

= b sin. A

b
also, in the right-angled triangle B C D we have-

P Р
sin. B Þ

= a sin. B

a hence it is evident that

a sin. B b sin. A, since each = Þ If the angle A is obtuse, then the perpendicular C D will fall without the triangle A B C, and B A must be produced to meet it; then in the right-angled triangle BCD we have

p
= sin. B

p = a sin. B

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but, since the angles D A C and A are supplemental their sines are equal. therefore,

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a

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Similarly, by drawing perpendiculars from the other angles it can be shown that

sin. A a:c = sin. A : sin. C; or

sin. C and

b sin. B 6:0= sin. B: sin. C; or

с

sin. C

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which is called the Rule of Sines, and may be expressed as follows

The sides of a plane triangle are in the same ratio as the sines of the opposite angles.

(1) A triangle is determined by one side and two angles; and inasmuch as two angles determine the third angle, there remains only to find the two sides.

(2) A triangle is determined (in all cases but one) by two sides and an angle opposite to one of them.

NOTE.—The doubt or ambiguity arises in the one case when the given angle is not that appertaining to the greater side. There is no ambiguity when the given angle is opposite to the greater side.

(I) To solve the triangle when one side and two angles are given, we have to find the remaining angle and the other two sides.

The remaining angle is found by adding together the two given angles and subtracting the sum from 180°, because the sum of the three angles of a plane triangle is always equal to 180°.

GENERAL RULE TO FIND A SIDE.-Write down the side required, and underneath it put the given side so as to form a fraction. Now put this fraction equal to its corresponding fraction in the Rule of Sines, and multiply both sides of the equation by the denominator of the first formed fraction; the resulting equation will give the formula by which the required side is found. To exemplify this, suppose the side b was given, and also the angles A and B, and we want to find the side a. Write down a, and underneath it put b, thus, ģ, the corresponding fraction in the Rule of

sin. A Sines is

Putting these fractions equal to one another, we have sin. B sin. A

then by multiplying by the denominator b, we have a = 5 sin. B'

sin. A ъх

Hence the side a will be found by adding together the logarithm sin. B of b and the sin. A, and from the sum subtracting the sin. B. The natural

number corresponding to the remainder will be the value of a. The side c will be found in a similar manner, but before it can be done the angle C must be found.

GENERAL RULE TO FIND AN ANGLE.—(2) To solve the triangle when two sides and an angle opposite one of them are given. Write down the sin. of the required angle (this must be always the angle opposite the other given side), and underneath it put the sin. of the given angle so as to form a fraction. Now put this fraction equal to its corresponding fraction in the Rule of Sines; the resulting equation will give the formula by which the required angle is found. To exemplify this, suppose the two sides a and b are given (of which a is greater than b) and the angle A. We must first find the angle B. Write down sin. B, and underneath it put sin. B

b. sin. A, thus, the corresponding fraction in the Rule of Sines is sin. A'

sin. B b Putting these fractions equal to one another, we have

sin. A

b and multiplying by the denominator sin. A we have sin. B x sin. A. Hence the angle B will be found by adding together the logarithm of b and sin. A, and from the sum subtracting the logarithm of a. The degree and minute corresponding to this remainder in the sin. column will be the value of the angle B. The other parts are found exactly as in (1).

a

a

a

Given one Side and two Angles, to find the other Sides, and the

remaining Angle.

Example. Given the angle A = 36° 15', the angle B = 105° 30', and the side A B = 53; required the sides A C and B C, and the angle C.

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Draw the line A B, and make it equal to 53; make the angle B AC 36° 15', and the angle A B C 105° 30', and draw the lines A C and BC till they meet in C; then A C will measure 82:5, and B C 50-62.

E 2

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