Sidebilder
PDF
ePub

Multiplying extremes and means, .0003319x=.00145800, and x=44" Therefore the angle belonging to the cotangent of 10.33554 is 24° 47' 16".

SOLUTIONS OF RIGHT ANGLED TRIANGLES.

CASE I. Given the hypothenuse and an angle, to find the base and perpendicular.

Example 2. See figure 14. To find the base AB. If the hypothenuse be made radius, BC will be the sine of the angle BAC, and AB sine of the angle ACB: then ac: AC:: ab AB, or R; AC::cos A: AB: By the table of natural sines, 1:250::0.6883546 172.088650=the base AB. By the table of logarithmic sines,

As radius 10.0000000

To hypothenuse 2.3979400-250

So is cos A 9.8378122-43° 30'
To the base 2.2357522-172.1.

To find the perpendicular BC. If the hypothenuse be made radius, R: AC::sin A: BC. By the table of natural sines, 1:250::0.7253744 181.348600 the perpendicular BC. By the table of logarithmic sines, &c.

As the radius 10.0000000

To the hypothenuse 2.3979400

So is the sin A 9.8605622

To the perpendicular 2.2585022=181.35.

Therefore the base is 172.1 rods, and the perpendicular

181.35.

Other proportions might be instituted for finding the base and perpendicular, by making different lines radius. If, for example, the base be made radius as in Fig. 15, the perpendicular will be tan, and the hypothenuse sec of A. Then sec AACR; AB; and, sec A: AC::tan A; BC. Also, if the perpendicular be made radius as in Fig. 16, the base will be tan and the hypothenuse sec of C. Then sec C AC::tan C: AB; and, sec C: AC::R BC.

Note. In the above solution, we have obtained the base and perpendicular by two methods, the table of natural sines, &c. and the table of logarithmic sines, &c. For future solutions in Trigonometry, one method will generally be considered sufficient; and, as operations by

logarithms are so much more expeditious than the mode of calculation by natural sines, it will in most cases be preferred.

CASE II. Given the hypothenuse and one leg, to find the angles and the other log.

Example 1. See Figure 14. Making the hypothenuse ra. dius, ACR::AB: sin C; that is, the hypothenuse is to radius, as the base is to the sine of the angle at C. But the hypothenuse 35, and the radius=26: therefore, 35: 1:: 26 sin C. By the tables of logarithmic sines, &c. As the hypothenuse 1.5440680

Is to radius 10.0000000

So is the base 1.4149733
To sin C 9.8709053

47° 59' 9.8709597

47° 58' 9.8708458

9.8709053

9.8708458

.00011391::.0000595.5=30".`

Therefore the angle belonging to sin C, is 47° 58′ 30′′. The other acute angle is equal to 90° diminished by 47° 58′ 30", which is 42° 1′ 30′′.

The perpendicular is found by Theorem I. R: AC: :sin A BC; and by the tables,

Radius 10.0000000

Is to the hypothenuse 1.5440680

As sin A 9.8258614

Is to the perpendicular 1.3699294

The natural number belonging to 1.3699294, is 23.439. Therefore the acute angles at C and A are 47° 58′ 30′′, and 42° 1' 30"; and the perpendicular is 23.439.

Example 2. See Figure 19. To find the angle at A, make hypothenuse radius, AC: R::BC: sin A; that is, by the tables,1.732394: 10.000000::1.681241 9.948847=62° 44'. Making the perpendicular radius, BC; R::AC; sec C; and, by the tables,

1.681241 10.000000::1.732394 .051153=27° 16'. To find the base, R; BC::tan C; AB; and, by the tables, 10.000000; 1.681241::9.7121461.393387=24.74. Therefore the angle at A is 62° 44′, and the base 24.74.

CASE III. Given the angles and one leg, to find the hy. pothenuse and the other leg.

Example 1. See Figure 20. To find the hypothenuse, make the hypothenuse radius; then sin C:AB::R; AC: and, by the tables,

9.832152: 1.778151;:10.000000 ; 1.945999=88.31. To find the perpendicular, R: AC:: sin A: BC; and, by the tables,

10.000000

1.945999;:9.865536

1.811535-64.8. Therefore the hypothenuse is 88.31, and the perpendicular 64.8.

Example 2. See Figure 21. To find the hypothenuse, make the hypothenuse radius; then sin A: BC::R: AC; and, by the tables,

9.6793602 1.8692317::10.000000: 2.1898715=154.83. To find the base, R; AC::sin C: AB; and, by the tables, 10.000000 2.1898715::9.9436925: 3.1335640, the nat ural number belonging to which is 136=AB.

Therefore the hypothenuse is 154.83, and the base is 136.

CASE IV. Given the base and perpendicular, to find the hypothenuse and the angles.

Example 1. See Figure 22. To find the angle belonging to the tan A, make the base radius; then AB¦R::BC; tan A; and, by the tables,

[ocr errors]

2.453318 10.000000::2.283301: 9.829983=34° 4′. The remaining acute angle is 90° diminished by 34° 4′, which equals 55° 56'.

To find the hypothenuse, R: AB:: sec A AC; and, by the tables,

10.000000 2.453318::10.081767: 2.535085, the natural number belonging to which is 342.84. Therefore the angle at A is 34° 4', and the hypothenuse AC is 842.84.

Example 2. See Figure 19. To find the angle belonging to the tan A, make the base radius; then AB; R::BC: tan A; and, by the tables,

2.806180 10.000000::2.6812419.874961=36° 52′ 12". The remaining acute angle at A is 90° diminished by 36° 52′ 12", which equals 53° 7' 48" To find the hypothenuse, R:AB::sec A: AC; and by the tables,

10.000000 2.806180::10.096910: 2.903090=800.

Therefore the acute angles are 36° 52′ 12′′, 53° 7′ 48′′, and the hypothenuse is 800.

Six examples under article 139.

Note. As the figures employed in the remaining solutions in this section, are nothing more than plain right angled triangles, where the lettering is uniformly the same, it is unimportant that reference should be made to any particular figure. The base is uniformly AB, the perpendicular BC, and the hypothenuse AC. The only advantage in employing one rather than another, is, that the proportions of the figure may be more correctly adapted to the conditions of the question, and even this could not, in all cases, be secured without drawing a figure entirely anew. In Application of Algebra to Geometry, the figures were so deficient that new ones have been furnished throughout; through the remaining part of the course, there will probably be little need of but few additions. They will, however, be furnished in every instance where they are wanting.

Example 1. CASE I. Make the hypothenuse radius: then to find the perpendicular, R: AC::sin A: BC; and, by the tables.

10.000000

1.832509::9.8015111-634020=43.06. To find the base, R: AC::sin C: AB; and by the tables, 10.000000 1.832509::9.888755 1.721264=52.64. Therefore the perpendicular is 43.06, and the base 52.64.

Example 2. By CASE II. Make the hypothenuse radius ; then to find the acute angles, ACR::AB: sin C; and, by the tables,

2.929419 10.000000::2.7737869.844367 = 44° 20′ nearly.

To find the perpendicular: By THEOREM I, R: AC: sin A: BC; and, by the tables, 10.000000 2.929419 :: 9.854480 2.783899-608.

[ocr errors]

Therefore the angle at C is 44° 20', and the perpendicular is 608. The remaining acute angle is 45° 40′.

Example 3. By CASE II. Make the hypothenuse radius: then, to find the angle at A, AC R::BC siu A; and, by the tables,

1.892095

10,000000::1.7558759.862780-46° 48′ 38′′. To find the base,R: AC::sin C: AB; and, by the tables, 10.000000 ; 1.892095: 9.835314; 1.727419=53.387.

Therefore the angle at A is 46° 48′ 38′′, and the base is 53.387.

Example 4. CASE III. Make the hypothenuse radius ; then, to find the hypothenuse, sin C:AB::R:AC; and, by the tables,

9.637184 2.859138::10.000000 3.221954=1667.4. To find the perpendicular, R AC: :sin A: BC; and, by the tables,

10.000000 3.221954: 9.954762: 3.176716=1502.2. Therefore the hypothenuse is 1667.4, and the perpendicular 1502.2.

Example 5. Make the hypothenuse radius. To find the hypothenuse, Sin A; BC::R; AC; and, by the tables,

9.995316 2.800717::10.000000 2.805401=638.9. To find the base, R: AC::sin C:AB; and by the tables, 10.000000 2.805434::9.164600 1.970034-93.44. Therefore the hypothenuse is 638.9, and the base 93.44.

Example 6. Solved by THEOREM II. CASE IV. Make the base radius. To find the angle A, AB R::BC: tan A; and by the tables,

1.5051500 52' 17".

10.0000000 :: 1.3802112

9.8750612=36°

To find the hypothenuse, RAB::sec A; AC; and by the tables,

10.0000000

1.5051500::10.0969105

1.6020605=40. Therefore the angle at A is 36° 52′ 11′′, and the hypothenuse is 40. The angle C-90°-36° 52′ 11′′ 53° 7′ 49′′.

Three examples under article 140.

Example 1. BC2+AB2=AC2; and by substituting numbers AC2=242+32=576+1024 1600, and by extracting the square root, AC=40=the hypothenuse.

Example 2. AC-AB2=BC2; and by substituting num. bers, BC2 = 1002-802-10000-6400=3600, and by ex. tractingthe square root, BC=60=the perpendicular.

Example 3. AC-BC-AB2: and by substituting

« ForrigeFortsett »