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In the calculation of right-angled plane triangles, any side, whether given or required, may be made radius to find a side; but a given side must always be made radius to find an angle. Hence, to find an angle, begin the proportion with a side opposite to a given angle; and to find a side, begin the proportion with an angle opposite to a given side.

69. This problem contains four particular cases, and the solutions, or rules of calculation, all depend upon the first and second propositions.

Case 1. When the hypothenuse and one of the acute angles are given, to find the rest.

2. When a side and one of the acute angles are given. 3. When the hypothenuse and one of the sides are given.

4. When the two sides are given.

In the following table the first column contains the things given, the second contains the things required, and the third contains the rules or proportions by which they are found.

GIVEN

SOUGHT

SOLUTION

I. AC & C, the hyp. & AB R : s. C :: AC : AB BCR cos. C:: AC: BC

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Remarks on the Solutions in the table.

70. In the second case, when the base AB and the angle A are given, to find the hypothenuse AC, a solution may also be obtained by help of the secant. For AB: AC :: R: sec. A, therefore R: sec. A :: AB: AC.

In the third case, when the hypothenuse AC and the side BC are given, to find AB, a solution may be obtained from 47. 1. For AB-AC-BC, therefore AB VAC2-BC2. This value of AB may be easily calculated by logarithms, if the quantity AC2-BC be resolved into two factors. Thus, AC2_BC2=(AC+ BC) × (AC—BC) (Cor. 5. 2); therefore AB=√(AC+BC) × (AC—BC). Hence, log. AB= log. (AC+BC) +log. (AC—BC)

2

In the fourth case, when AB and BC are given, AC may be found from 47. 1. For AC=√AB2 + BC2. But AB2 + BC2 cannot be separated into multipliers; therefore when AB and BC are represented by great numbers, this rule is not conve nient for computation by logarithms. It is then best to seek first the tangent of A, by the analogy in the table, AB: BC:: R: tan. A. But if the angle A be not required, it is sufficient, after tan. A is found by this proportion, to take from the trigonometrical tables the cosine which corresponds to tan. A, and then to compute AC by this proportion, cos. A:: R:: AB: AC.

71. If the two acute angles of a right-angled triangle be denoted by A, C, and their opposite sides by a, c respectively, and the hypothenuse by b; then prop. I, II may be expressed by general equations, by means of which all the cases of rightangled triangles may be easily resolved.

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2.

c tan. A

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or log. a=log. c + log. tan. A—log. r.

If any two of these three parts (beside radius) be given, the third part may be found from these equations.

PROBLEM II.

72. In any oblique-angled triangle, of the three sides and the three angles three parts being given, and one of these being a side, it is required to find the other three parts.

This problem contains four cases, and the solutions of them depend upon the preceding propositions.

Case 1. When two angles and a side are given, to find the rest.

2. When two sides and an angle opposite to one of them are given.

3. When two sides and the included angle are given. 4. When the three sides are given.

73. The following properties of plane triangles are requisite in the solution of the cases of oblique-angled triangles.

The sum of the three angles of every plane triangle is equal to two right angles, or 180 degrees (32. 1). Hence

1. When two angles of a triangle are given, the third angle is said to be given; for it is the supplement of the other two angles, and is found by subtracting their sum from 180°.

2. When one angle of a triangle is given, the sum of the other two angles may be found by subtracting the given angle from 180°.

The angles opposite to the two least sides of a plane triangle are acute; and if there be an obtuse angle, it is opposite to the greatest side.

A perpendicular drawn from the opposite angle to the longest side will fall within the triangle; and the greater segment will be next to the greater side, and the less segment to the less side.

If both the angles at the base be acute, the perpendicular will fall within the triangle; but if one of them be obtuse, the perpendicular will fall without the triangle, beyond the obtuse angle; and in both cases the greater segment will be next to the greater side, and the less segment to the less side.

74. Solution of the Cases of Oblique-angled Triangles.

Case 1. Given two angles A and B, and one side AB, to find the other two sides. See fig. art. 63.

Solution. Because the angles A, B are given, the angle C is also given, being the supplement of A+B (73).

Then

sine C: s. A :: AB : BC (54),

and sine C: s. B:: AB : AC.

Case 2. Given two sides AB, AC, and the angle B opposite to one of them, to find the other angles Ă, C, and the third side BC.

Solution. AC: AB :: s. B: s. C (54). Hence ▲A= 180°—B—C. Then, s. B: s. A :: AC : CB.

Case 3. Given two sides AB, AC, and the angle A contained between them, to find the other angles B and C, and the third side BC.

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Solution. AB+ AC : AB—AC :: tan. (B+C): tan. (B-C) (61). Since (B+C) and (B-C) are given, the angles B and C may be found. For B= (B+C)+ (B-C) (57), and C=1 (B+C)- (B-C). Then s. B: s. A :: AC: BC.

Remark. Instead of the tangent of half the sum of the two unknown angles we may take the cotangent of half the given angle, or the tangent of half its supplement; for these three tangents are equal to one another. Thus, the sum of the angles B and C is less than two right angles (32. 1), therefore half their sum is less than one right angle, therefore tan. (B+C) cot. their complement A (23). Also, 180°- AB+ C. Hence the first analogy becomes, AB+ AC: AB-AC :: cot. A or tan. (180°-A): tan. (BC).

Otherwise. Given AB, BC, and the included angle B, to find the rest.

Upon one of the given sides, as BC, let fall a perpendicular AD from the opposite angle A. Then, in the right-angled triangle ABD the hyp. AB and all the angles are given; therefore R: cos. B:: AB: BD (49). Now BD being found, and BC being given, DC will be given. Hence BD : DC :: tan. given angle DAB : tan. DAC (53). Hence BAC=BAD+DAC, when the perp. falls within the triangle, and BAC-BAD-DAC, when the perp. falls without. Hence C-180°-B— BAC. Lastly, s. C: s. B:: AB : AC.

Remark. When two sides and the included angle of a plane triangle are given, the third side may be found by art. 64, without first finding the angles. But this method of finding the third side is not adapted to calculation by logarithms, because the quantity under the radical sign cannot be resolved into simple factors. Therefore the operation by this rule will be tedious, unless the two given sides be expressed by small numbers. When the two given sides are expressed by great numbers, it will be more convenient to find the angles and the third side by either of the preceding methods.

Case 4. Given all the sides, to find the angles..

Solution. Make the longest side the base, and let fall a perp. upon it from the opposite angle. The angles B and C, adjacent to the longest side, will be acute. Let F denote the difference of the segments of the base. Then BC: BA+AC :: BA-AC: F (59). Hence (BC+F)= greater segment of the base (57), and (BC-F)=less segment. Then CA: CD: R: cos. C, and BA: BD :: R cos. B. Hence A=180°-B-C.

When the sides are not expressed by great numbers, the angles may be easily found by art. 65.

Cos. B=

AB +(BC+AC)×(BC—AC)

2 AB. BC

D

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