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Here the arc ab cuts AC in one point only, because AB is greater than BC; therefore the angle A is acute, and not ambiguous.

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To the angle C = 33° 21', add the angle A = 26° 21', and the sum is 59° 42, which subtracted from 180°, leaves the angle B = 120° 18'.

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By Gunter's Scale. 1. Extend the compasses from .98 to .79 on the line of numbers, that extent will reach from 33° 21' to 26° 21', the anglè A, on the line of sines.

33° 21',

2. Add the angle A = 26° 21' to the angle C and the sum will be 59° 42'; then extend the compasses from 33° 21' to 59° 42', on the line of sines, that extent will reach from .98 to 1.54, the side AC, on the line of

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2. In the triangle ABC, are given the angle O = 33° 21', the side BC = 95.12 and the side AB = 60, to find the angles A and B, and the side AC.

By Construction, Fig. 44 This is constructed in the same manner as the preceding example; only AB, being shorter than BC, the arc ab cuts AC in two points on the same side of BC; hence the angle A may be either acute or obtuse. The side required has also two values as AC and AÇ.

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The sum of the angles C and A subtracted from 180° leaves the angle B = 86° 1' if A be acute, or 27° 17' if A be obtuse.

To find the side AC answering to the acute valne of the angle A. As sine of C, 33° 21'

9.74017

Is to sine of B, 86° 1'
So is AB, 60

9.99895 1.77815

11.77710

To find the side AC, answering to the obtuse value of the angle A.

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3. In a triangle ABC, the side AB is 274, AC 306, and the angle B 78° 13'; required the angles A and C, and the side BC. Ans. A = 40° 33', C = 61° 14', and BC = 203.2.

4. In a right angled triangle, there are given the hy. pothenuse AC = 272, and the base AB = 232; to find the angles A and C, and the perpendicular BC. Ans. A = 31° 28, C 58° 32' and BC = 142.

5. In a right angled triangle ABC, the hypothenuse AC is 150*and one side BC 69; required the angles and other side. Ans. O = 62° 37', A

62° 37', A = 27° 23' and AB 133.2.

CASE 3.

Two sides and the included angle being given, to find

the other angles and side.

RULE.

Subtract the given angle from 180°, and the remain.

As the sum of the two given sides,
Is to their difference;
So is the tangent' of half the sum of the two un-

known angles,
To the tangent of half their difference.*

This half difference of the two unknowo angles, added to their half sum, will give the angle opposite the greater of the two given sides, and being subtracted from the half sum, will give the angle opposite the less, given side.

After finding the angles, the other side may be found by Case 1.

• DEMONSTRATION. Let ABC, Fig. 45, be the proposed triangle, having the two given sides 1B, AC, including the given angle A. About A as a centre, with AC the greater of the given sides, for a distance, describe a circle meeting AB produced in E and F, and BC in D); join DA, EC, and FC, and draw FG parallel to BC, meeting EC in G.

The angle EAC (32. 1.) is equal to the sum of the unknown angles ABC, ACB, and the angle EFC at the circumference is equal to the half of EAC at the centre (20.3.); therefore EFC is hálf the sum of the unknown angles; but (32 1.) the angle ABC is equal to the sum of the angles BAD and ADB or BAD and ACB; therefore FAD is the difference of the unknown angles ABC, ACB, and FCD, at the circumference is the half of that difference ; but because of the parallels DC, FG, the angles GFC, FCD are equal, therefore GFC is equal to half the difference of the unknown angles ABC, ACB; but since the angle ECF in a semicircle, is a right angle, EG is perpendicular to CF, and therefore CF being radius, EC, CG are the tangents of the angles EFC, CFG; it is also evident that EB is the sum of the sides BA, AC, and that BF is their difference; therefore since BC, FG are parallel EB : BF :: EC : CG (2. 6.); that is, the sum of the sides AC, AB, is to their difference, as the tangent of half the sum of the angles ABC, ACB, is to the tangent of half their difference.

To demonstrate the latter part of the rule, let AC and AB, Fig. 46, represent any two magnitudes whatever; in AB produced take BD equal to AC the less, and bisect AD in E.

Then because AE is equal to ED and AC to BD, CE is equal to EB; therefore AE or ED is half the sum of the given magnitudes AB, AC, and CE or EB is half their difference; but AB the greater is equal to AE, EB, that is to half the sum added to half the difference, and AC the less, is equal to the excess of

EXAMPLES.

1. In the triangle ABC, there are given AB = 1280 AC 90, and the angle A = 49° 12, to find the angles B and C, and the side BC.

By Construction, Fig. 47.

Draw AB = 128, and make the angle A = 48° 12; draw AC = 90, and join BC. The angle B will measure 44° 37', the angle C 87° 11', and the side BC 95,5

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180° 0 Angle A

48 12 Sum of the angles B and C 131 48 Half sum

do.

65 51

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As the sum of the sides AB, AC, 218

2.33846 Is to their difference, 38

1.57978 So is the tang. of half sum of angl. B &C, 65° 54' 10.34938

11.92916

To tang. of half their difference, 21° 17'

9.59070

Half sum of the angles B and C 65° 54
Add and subtract half their difference 21 17

Angle C

87 11

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