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Plate V.

A 46° 30′
B 37. 30

180°-84.00-96° 00-C.

By def. 29. sect. 1. The sine of 96°- the sine of 84°, which is the supplement thereof; therefore, instead of the sine of 96°, look in the tables for the sine of 84°.

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Extend from 84° (which is the supplement of 96°) to 46° 30′ on the sines; that distance will reach from 230 to 168, on the line of numbers, for BC.

Plate V.

Extend from 84° to 37° 30', on the sines; that extent will reach from 230 to 141, on the line of numbers, for AC.

CASE III.

Two sides and a contained angle given; to find the other angles and side.

In the triangle ABC, there is AB 240, the angle A 36° 40′ and AC 180, given; to find the angles C and B, and the side BC. Fig. 16.

Geometrically.

Draw a blank line, on which from a scale of equal parts, lay AB 240; at the point A of the line AB, make an angle of 36° 40′, by a blank line; on which from A, lay AC 180, from the same scale of equal parts; measure the angles C and B, and the side BC, as before; and you have the answer required.

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By Calculation.

By cor. 1. theo. 5. sec. 1. 180-the angle A 36°. 40' 143°. 20′ the sum of the angles C and B: therefore half of 143°. 20', will be half the sum of the two required angles, C and B.

By theo. 2. of this sect.

As the sum of the two sides AB and BC is to their difference,

420 60

Plate V.

So is the tangent of half the sum of
the two unknown angles C and B
to the tangent of half their difference

By Theo. 4.

71° 40' 28° 20′

To half the sum of the angles C and B
Add half their difference as now found

71° 40′ 23 20

The sum is the greatest angle or ang. C. 95 00

Subtract and you have the least angle, or B 48 20

The angle C and B being found; BC is had, as before, by theo. 1. of this sect. Thus,

S. B: AC:: S. A : BC.
48° 20′ 180 36°, 40 143. 9.

By Gunter's scale.

Because the two first terms are of the same kind, extend from 420 to 60 on the line of numbers; lay that extent from 45° on the line of tangents, and keeping the left leg of your compasses fixed, move the right leg to 71°. 40'; that distance laid from 45° on the same line will reach to 23° 30', the half difference of the required angles. Whence the angles are obtained, as before.

The second proposition may be easily extended, from what has been already said.

Plate V.

CASE IV.

The sides given, to find the angles.

In the triangle ABC, there is given, AB 64, 7 AC 74, BC 34: the angles A, B, C, are required.

fig. 17.

Geometrically.

The construction hereof must be manifest, from prob. 1. sect. 1.

By Calculation.

From the point C, let fall the perpendicular CD on the base AB; it will divide the triangle into two right-angled ones, ADC and CBD; as well as the base AB, into the two segments, AD and DB.

AC 47

BC 34

Sum 81

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Difference 13

By theo. 3. of this sect.

As the base of the longest side, AB

64

is to the sum of the other sides, AC and BC, 81

So is the difference of those sides

13

to the difference of the segments of
the base AD, DB.

16.46

Plate V.

By theo. 4. of this sect.

To half the base, or to half the sum
of the segments AD and DB
Add half their difference, now found,

32

8.23

Their sum will be the greatest segment AD 40.23

Subtract, and their difference will be the least segment DB,

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23.77

In the right angled triangle ADC, there is AC 47, and AD 40. 23, given to find the angle A.

This is resolved by case 4. right-angled plane trigonometry, thus,

ADR AC: Sec. A.

40.23 : 90°. : : 47: 31° 08′

Or it may be had by finding the angle ACD, the complement of the angle A: without a secant, thus,

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90-58° 52′ 31° 08′ the angle A.

Then by theo. 1. of this sect.

BC. S. A:: AC: S. B.
34: 31° 08′:: 47: 45° 37'.

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