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PART II. FUNDAMENTAL PRINCIPLES OF SOLUTION

19. General Method for Oblique Triangles. In the solution of oblique triangles the following cases arise:

Case I.

Case II.

Case III.

Given two angles and a side.

Given two sides and the included angle.

Given the three sides.

Case IV. Given two sides and an angle opposite one of them. A general method for solving oblique triangles in all of these cases consists in dividing the triangle into two right triangles by a perpendicular from a vertex to the opposite side; these right triangles are then solved by the methods of the previous chapter. In all cases except the three side case the perpendicular can be drawn in such a manner that one of the resulting right triangles contains two of the given parts.

This method applied in the various cases leads to formulas for the solution if letters are employed for the sides and angles.

20. Case I: Given Two Angles and a Side. In this case it is immaterial which side is given, since the third angle can be found immediately from the fact that the sum of the three angles is 180°. Drop the perpendicular from either extremity of the given side.

De

Example 1. An oblique triangle has one angle equal to 43°, another equal to 67°, and the side opposite the unknown angle equal to 51. termine the remaining parts.

B

43

51

It is immediately seen that the third angle is 180° — (43° + 67°) = 70°. To solve this triangle draw the figure approximately to scale and drop the perpendicular CD = p from one extremity C of the known side to AB, the side opposite C. Denote the unknown side CB by a. In the right triangle ACD, the hypotenuse and one angle are known; hence by (13), § 6, p = 51 sin 67° 46.95. An angle and the side opposite, in the right triangle BCD, are now known; hence by (15), § 6, a p/sin 70°

=

=

=

46.95/.9397 49.96.

FIG. 21

The side AB may be found in the same manner. Check as in § 3, p. 2.

21. Case II: Given Two Sides and the Included Angle. The triangle can be divided into two right triangles, one of which contains two of the known parts, by a perpendicular drawn from either extremity of the unknown side to the side opposite. Two sides of a triangle are 26.5 and 32.8; the included Find the remaining parts.

Example 1. angle is 58° 18'.

In the figure let AB = 32.8, AC = 26.5, and the angle at A= 52° 18'. Drop a perpendicular p from B to the opposite side. Denote the unknown side by a and the segments of AC by x and

B

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y as in Fig. 22; then p, x, y, a, can be computed in the following order:

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The student may determine the angles at B and C.

22. Case III: Given the Three Sides. In this case it is not possible to divide the triangle into two right triangles in such a way that one of them contains two of the given parts; however, if a perpendicular is dropped to the longest side from the vertex of the angle opposite, the segments into which this side is divided by the perpendicular are easily computed, as in the example below. There is one and only one solution, provided no side is greater than the sum of the other two.

Example 1. The sides of a triangle are a = 36.4, b = 50.8, and c = 72.5. Determine the angles.

Draw a figure and drop a perpendicular from B upon AC. Denote the segments of the base by x and y as in Fig. 23; then

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Since we now know x and y, the angles at A and C are easily found. The student may complete the solution. See also § 25.

23. Case IV: Given Two Sides and the Angle Opposite One of Them. The triangle is easily solved by the general method, dropping the perpendicular from the vertex of the angle included by the given sides.

Example 1. One angle of a triangle is 37° 20'; one side adjacent is 25.8 and the side opposite is 20.8. Solve the triangle.

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First construct the given angle A and on one side of A lay off AB = 25.8. With B as center and radius = 20.8 describe an arc of a circle meeting the opposite side in two points C and C'. Either of the triangles ABC, ABC' satisfies the given conditions; the case is on this account called the ambiguous case.

The student should note that the triangle BCC' is isosceles and that the interior angle of ABC at C is equal to the exterior angle of ABC' at C'; hence the interior angles C and C' are supplementary. To solve ABC draw the perpendicular BD = p from B; then determine p from the right triangle ABD.

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25-8

87. 20

20.8

FIG. 24

20.8=a>p

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hence C is the acute angle whose sine is .75223; i.e. C = 48° 47′.05. The student can complete the solution as follows:

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C'′ = 180° - C; B' = 180° — (A+C'); AC' = AD — CD.

EXERCISES X.-SOLUTION OF TRIANGLES

Find the remaining parts of the following triangles by suitably dividing each into two right triangles. Capital letters represent angles; small letters the sides opposite them.

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To determine the distance from a point A to an inaccessible object

B, a base line AC 300 ft. and the angles BAC = 40°, BCA = 50° are

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6. To determine the distance between two trees A, B on opposite sides of a hill, a point C is chosen from which both trees are visible; the distances AC 400 ft., BC 361 ft., and the angle ACB = 55° are then measured. What is the distance between the trees?

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7. The sides of a triangular field are 43 rods, 48 rods, and 57 rods, respectively; determine the angles between the sides.

8. A 50 ft. chord of a circle subtends an angle of 100° at the center. A triangle is to be inscribed in the larger segment having one of its sides 40 ft. long. How long is the other side? Is there only one solution?

9. A triangle having one of its sides 60 ft. long is to be inscribed in the segment of Ex. 8. Determine the remaining side. How many solutions are there in this case ?

24. The Law of Sines. In Example 1, § 20, it may be observed that it was not really necessary to calculate p numerically in order to find a, for we might have written

FIG. 25 a

P=b_sin_A

p = a sin B

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from which a could have been found by the use of logarithms in one calculation.

A formula which can be used B for all examples of this case can be obtained in a similar manner.

Let us denote the sides opposite

A, B, C, by a, b, c, respectively. The adjoining figures represent two cases: in one the triangle has all its angles acute; in the other, one angle,

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or, dividing first by the product sin A sin B, and second by

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If the perpendicular is drawn from one of the other vertices, say from B, the above procedure leads to the equation

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This result, known as the law of sines, may be stated as follows:

In any triangle any two sides are proportional to the sines of the angles opposite them.

By using this law any example under Case I can be solved as in the following example. There is always one and only one solution if the sum of the given angles is less than 180°.

Example 1. Given one side of a triangle a = 2.903 and two of the angles B = 79° 40′, C′ = 33° 15'; find the remaining parts.

We first obtain the angle A opposite the given side a and then apply the law of sines.

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* The fact that a proportion x/a = b/c gives x = ab/c should be memorized by some device. Then log x = log a + log b+colog c.

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