Sin. G: R:: HK: KG, and tan. G: R:: KH : HG. HK 186+ R. log. 12-2695129 HK+R. log. 12-2695129 Sin. G 43° 38′ log. 9-8388747 tan. G log. 9.9792738

GK 269-549 log. 2.4306382 GH 195-09 log.

4. Given the hypotenuse LN 415 inches, and the perpendicular MN 249; to find the angles, and LM.

LN: NM:: R : sin. L.

2.2902391

N

M

NOTE. LM is equal to the square root of the product of the sum and difference of LN and NM√664 × 166 =

perpen

T

5. Given the base RS 53 miles, and the dicular ST 67; to find the angles, and hypotenuse

RS: ST:: R: tan. R.

RT.

ST 67+ R.

RS 53

Tan. R 51° 39' 16"

log. 11.8260748 R

log. 1-7242759

log. 10.1017989

R: sec. R:: SR: RT.

Sec. R 51° 39′ 16′′. Rad. log. 0-2073261

RS 53

RT 85.4284

log. 1.7242759

log. 1.9316020

NOTE. The square of RT is equal to the sum of the squares of RS and ST; therefore TR√53o +67o = √729885·4284.

6. Given the hypotenuse 893, and the base 586 chains. Ans. Angle at base 48° 59′ 17′′, perpendicular 673-832 ch. 7. Given the base 326 yards, and the vertical angle 64° 40'. Ans. Hypotenuse 360-686, perpendicular 154.33 yards. 8. Given the perpendicular 286, and vertical angle 71° 24'. Ans. Hypotenuse 896-666, base 849-8329. Given the hypotenuse 963 links, and vertical angle Ans. Base 641.87, perpendicular 717-89 links

41° 48'.

OBLIQUE TRIANGLES.

IF two angles of a triangle be known, the third is got by subtracting their sum from 180°; and if one angle be known, the sum of the other two is got by subtracting it from 180o.

RULE I. Any two sides of a triangle are to one another as the sines of the angles opposite to them. Thus BC: CA :: sin. A sin. B, or sin. A: sin. B:: CB: CA. The former order is to be used when an angle is required, and the latter when a side.

NOTE. This rule is to be used whenever a given angle is opposite to a given side.

The

1. Given two sides AB 532, and BC 358 feet, and the angle at C 107° 40′; to find the angles at A and B, and AC. figure is drawn by Prob. XX. PRACTICAL GEOMETRY.

AB: BC:: sin. C: sin. A.

B

Д

Sin. C (107° 40') 72° 20'* log. 9.9790192
BC 358 feet

BA 532

Sin. A 39° 53'

log. 2.5538830

12.5329022

log. 2.7259116

log. 9.8069906

B=180° (C+A), and sin. C: sin. B:: BA: AC.

-

Sin. B 32° 27'

BA 532

Sin. C 107° 40'

AC 299.6

2. Given AB 232, and BC 345 yards, and the angle at C 37° 20'.

By proceeding in the same way, the angle at A may be either 64° 24' or 115° 36', and therefore the angle at B may be either 78° 16′ or 27° 4′, and AC 374-56 or 174-07. For AB being less than BC, there are two triangles which have each of them the given things in them.

3. Two places are 560 feet from one another, and at a station 258 feet from the first place, their distance subtended an angle of 63° 28'. Required the distance of the station from the other. Ans. 625-469 feet.

• When the angle is greater than 90°, take the sine, tangent, &c. of its supplement.

4. Given two angles D 63° 48′, and E 49° 25′,
and the side EF opposite to D 275 yards; to find
DE and DF. Constructed by Prob. XXI. PRAC-
TICAL GEOMETRY. The angle at F is
-(D+E)=66° 47'.

Sin. D: sin. E:: EF: FD.
Sin. E 49° 25′ log. 9.8805052

180°

F

5. Given the angles at E 49° 25′, and F 63° 48′, and the side EF 275; to find ED and DF.

Ans. ED 268-488, and DF 227.2546,

6. A ship sailing due north observes a cape bearing N. 54° 12′ W.; and after sailing 27 miles, the cape bore S. 70° 30′ W. Required her distances from it.

Ans. First distance 30.957, second distance 26-636 miles.

RULE II. When two sides and the angle between them are given.

Add and subtract the sides to get their sum and difference. Subtract the angle from 180°, and take half the remainder, to get half the sum of the unknown angles. Then as the sum of the sides is to their difference, so is the tangent of half the sum of the unknown angles to the tangent of half their difference. Having thus found the half difference, add it to the half sum to get the angle opposite to the greater side, and subtract it to get the less angle; after which the third side is found by Rule I.

7. Given the sides GH 133, and HK 176 yards, and the angle at H 73° 16′; to find the angles at G and K, and the side GK.

KH+HG: KH-HG:: tan. (G+K): tan. (G-K).

KH-HG 43

log. 1.6334685

Tan. (180°-H) 53° 22′ log. 10-1286790

8. Given GH 237, and GK 482 feet, and the angle at G 77° 48'; to find the angles at H and K, and HK.

Ans. H 73° 59′ 39′′, K 28° 12′ 21′′, and HK 490.1144 feet. 9. Given HK 78, and KG 168, and the angle K 128° 26'. Ans. H 35° 48′ 20′′, G 15° 45′ 40′′, HG 224-94.

RULE III. When the three sides are given.

Add the three sides, and from half the sum subtract the side opposite to the angle sought; then take the arithmetical complements of the two sides containing the angle sought, and the logarithms of the half sum and of the remainder, and add these four together, and half the sum will be the cosine of half the angle sought.

10. Given the sides SP 230, PR 365, and SR 426 feet; to find the angles.

SP 230 ar. co. 76382722

PR 365 ar. co. 7.4377071
SR 426

1021

Sum 510-5 log. 2.7079957
426

Rem. 84-5 log. 1.9268567

)19-7108317

IP 44° 12' 24" cosine 9.8554158

P 88° 24' 48"

In the same manner the angle S is 58° 55′ 25′′.

11. Given the sides SP 1248, PR 728, and RS 956 feet. Ans. The angle R 94° 40′ 50′′, P 49° 46′ 16′′.

12. Given SP 375, PR 275, and RS 196.

Ans. The angle S 45° 17′ 26′′, P 30° 25′ 57′′.

PROMISCUOUS EXAMPLES.

1. Given the hypotenuse of a right-angled triangle 528 feet, and one of the acute angles 39° 27'.

Ans. The opposite side 335.57, adjacent side 407-7 feet. 2. Given the base 256, and the adjacent angle 57° 28'.

Ans. Hypotenuse 476-022, perpendicular 401-324 feet. 3. Given the perpendicular 297 feet, and the angle at the base 36° 48'. Ans. Hypotenuse 495'8, base 397 feet. 4. Given the hypotenuse 1268, and perpendicular 428 yards.

Ans. The base 1193-583, adjacent angle 19° 43′ 37.3". 5. Given the base 674, and the perpendicular 438 yards. Ans. Hypotenuse 803-816 yards, angle at base 33° 1' 4". 6. Given the hypotenuse 97, and the base 38 miles. Ans. Perpendicular 89-247 miles, angle at base 66° 56' 11". 7. Given the base 326, and the vertical angle 67° 30′.

Ans. The hypotenuse 352-86, perpendicular 135-034. 8. In an oblique triangle, given two angles 46° 48′ and 114° 26', and the side opposite the lesser 254 feet.

Ans. Other sides 317.233 and 112.097 feet. 9. Given two angles 56° 24′ and 74° 28′, and the side between them 354. Ans. Other sides 451 011 and 389-898. 10. Given two sides 572 and 748, and the angle opposite to the greater 67° 30'.

Ans. Angle opposite less 44° 57' 2", third side 748-267. 11. Given two sides 356 and 294, and the angle opposite to the lesser 51° 27'.

Ans. Other angles 71° 15′ 38.2" and 57° 17′ 218", or 108° 44′ 21.8" and 19° 48′ 38.2"; third side 316-31 or 127.407.

12. Given two sides 1864 and 1235, and included angle 73° 38'.

Ans. Other angles 68° 21′ 15.4′′ and 38° 0′ 44.6", third side 1924.2.

13. Given two sides 436 and 219, and included angle 127°. Ans. Other angles 35° 52′ 45.7" and 17° 7′ 14.3", third side 594-15.

14. Given the three sides 456, 327, and 184 yards.

Ans. Angles 123° 55′ 10′6′′, 36° 31' 3.2", and 19° 33′ 46.2". 15. Given the sides 2586, 1482, and 1234.

Ans. Angles 144° 14′ 53", 19° 33′ 47′′, and 16° 11′ 20′′.