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3. Making BC the radius.

T.C: AB :: Sec. C: AC.

T.C: AB :: R: BC. That is, as the tangent of C=47° 40 is to AB,

190 So is the Secant of C=47° 40'

10.040484

2.278754 10.171699

12.450453

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Or, having found one of the required sides, the other may be obtained, by one, or the other of the cors. to theo. 14. sect. 4.

3d. By Gunter's Scale:

1. When AC is made the radius.

Extend from 47° 40', to 90° on the line of sines that distance will reach from 190 to 257, on the line of numbers, for AC.

2. When AB is made the radius, the first stating is thus performed :

Extend from 45° on the tangents (for the tangent of 45° is equal to the radius, or to the sine of 90° as before) to 42° 20'; that extent will reach from 190, on the line of numbers, to 173, for BC.

3. When BC is made the radius, the second stating is thus performed:

Extend from 47° 40' on the line of tangents, to 45', or radius ; that extent will reach from 190 to 173, on the line of numbers, for BC; for the tangent of 47° 40', is more than the radius, therefore the fourth number must be less than the second, as before,

The two first statings of this case, answer the question without a secant.

4th. Solution by Natural Sines.

= BC.

ABXR.

AB*S of A
= AC; and
S of C.

S of C.
Nat. S of C, side ABXR.
Thus .739239) 190.000000 (257.02 &c.=AC.

147.8478

4215220
3696195

5190250
5174673

1557700
1478478

and,
.673443=Nat. S. of A.

190=side AB.

60609870

Nat. S. of C. 673443

.739239) 127.954170 (173.09=BC.

739239

5403027
5174673

2283540
2217717

6502300
6653151

CASE III.

The angles and perpendicular given; to find the base and

hypothenuse.

Pl. 5. fig. 6.

In the triangle ABC, there is the angle A 40°, and consequently the angle C 50°, with BC 170, given : to find AC and AB.

1st. By Construction.

Make an angle CAB of 40° in blank lines; (by prob. 16. sect. 4.) with BC 170, from a line of equal parts draw the lines EF parallel to AB (by prob. 8. sect. 4.) the lower line of the angle, and from the point where it cuts the other line in C, let fall a perpendicular BC (by prob. 7. sect 4.) and the triangle is constructed: the measures of AC and AB, from the same scale that BC was taken, will answer the question.

What has been said in the two foregoing cases, is sufficient to render the operations in this, both by calculation, Gunter's scale, and Natural sines, so obvious, that it is needless to insert them ; however, for the sake of the learner, we give for

Answers ; AC 264. 5, and AB 202. 6.

CASE IV.

The base and hypothenuse given ; to find the angles and per:

pendicular.

PL. 5. fig. 7.

In the triangle ABC, there is given, AB 300 and AC 500 : the angles A and C, and the pers pendicular BC, are required.

ist. By Construction.

From a scale of equal parts lay 300 from A to B; on B erect an indefinite blank perpendicular line, with AC 500, from the same scale, and one foot of the compass, in A, cross the perpendicular line in C; and the triangle is constructed.

By prob. 17. sect. 4. measure the angle A, and let Bd be measured from the same scale of equal parts that AC and AB were taken from; and the answers are obtained.

2d. By Calculation.

1. Making AC the radius.

AC:R:: AB: S.C.
R: AC:: 8.4.: BC.

Q

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By cor, 2. theo. 5. 90°- 36° 52' = 53° 08' the angle A. As radius

90°

10.000000 is to AC,

500

2.698970 So is the sine of A=53° 08'

9.903108

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Or BC may be found from cor. 2. theo. 14. sect. 4.

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