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Therefore instead of the proportion sin B : sin C:: AC: AB, we may always substitute sin B : cos B :: AC : AB. But since (10)... sin B : cos B :: lang B : R, we have

AC: AB : tang B: R:: cot C:R These principles suffice to solve all the cases of the right-angled triangle ABC, when besides the right-angle A, we know any two of the five parts B, C, AB, AC, BC; provided they are not the two angles. In this latter case we can only determine the ratios of the three sides. See the following

TABLE
For the solution of right-angled Triangles.
Given. | Required. |

Formulas.
AB, AC

BC BC=N AB? + AC?

B AB : BC::R: tang B

C AC: AB :: R: tang C
AB, BC

AC AC = VBC2—AB?

B BC: AB:: R: cos B

С BC : AB::R: sin C
AC, BC

AB AB=BC-AC2
B BC : AC::R: sin B

C BC : AC::R: cos C
AB, B

AC

R : tang B:: AB: AC

BC cos B:R AB : BC
AC, B

AB

R: cot B:: AC: AB

BC sin B: R:: AC: BC
AB, C

AC R: cot C;: AB; AC

BC sin C:R:: AB : BC
AC, C

AB

R; tang C::AC: AB

BC cos C:R::AC: BC
BC, B

AB R: : cos B:: BC: AB

AC R: sin B :: BC: AC
BC, C

AB R sin C:: BC : AB
AC

R: cos C:: BC : AC

Examples of the solution of right-angled Triangles. Ex. 1. In the right-angled plane triangle ABC, given the base AB=140, and the hypothenuse BC=335, required the perpendicular AC, and the angles B and C.

By Euclid (47. 1) and (38) of the present article, or by the preceding table we have

AC=(BCP-AB)=x (335-140)=304. 34 and BC : AB :: R: : cos B, whence RX AB 140

3.41791 BC 335 and by inspecting the table of natural sinos and cosines at the end

cos B =

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of this work, we shall find that the angle whose cosine=417913 to radius 1 is 65° 18'; consequently the other acute angle=90° 65° 18'—24° 42'.

But in working trigonometrical examples, it is better to avail ourselves of the logarithmic tables, as in most cases they enable us to dispense with the tedious multiplications and divisions which commonly result from employing the natnral sines and tangents of angles and arcs. Thus in the above example, since AC=V BC-AB”, log AC=

1 log VBC_AB'=

2

2{log (BC+AB) +log(BC—AB)}= {log _475+log 195}

= 10.

But log 475=2.676694 log 195=2.290035

2 4.966729 log AC=2.483364 whence AC=304. 34

RX AB Again cos B=

and therefore log cos B=log R+log AB

BC -log BC. But log R...... log AB (140) 2.146128

12.146128 log BC (335) = 2.525045 log cos B

9.621083 Whence the angle B=65° 18', and therefore < C=24° 42', as before.

EXAMPLE 2. In the right-angled plane triangle ABC (eame figure) given BC=33.249, and the angle B=17° 12' 51"; reqaired the base AB, the perpendicular AC, and the angles B and C. By (38) or by table R: cos B :: BC : AB

cos Bx BC

BC

R
R: sin B :: BC: AC (=

sin B x BC

B BC)

R But log cos B (17° 12' 51")= 9.980097 log BC (33. 249) = 1.521784

11.501881 log radius

10. log AB

= 1.501881 and AB=31. 76

Again to find AC

= 10.

= Io.

og sln B (17° 12' 51") = 9.471209
log BC (83. 249) = 1.521783

10.992992
log radius
log AC

0.992990 whence AC=9.84

EXAMPLE 3. In the right-angled triangle ABC (same figure) given the perpendicular AC=43, and the base AB=55, required the hypothenuse BC, and the angles B and C

By Euclid (47.1), or by table BC=VAB? + AC?=552 + 432 -69.81

RX AC and by (38) or by table, tang B

AB But log radius log AC (43) = 1.633469

11.633469 log AB (55) = 1.740363

log tang B = 9.893106 whence Z B=38° 1' 8", and LC=519 58' 52"

Instead of subtracting a logarithm it is usual to employ the arithmetical complement of that logarithm, which enables us to add at once the three terms which enter the proportion, without interrupte ing the operation on account of the subtraction.

The arithmetical complement of a logarithm is found from the tabular logarithm, by subtracting the right-hand figure of this latter from ten, and every one of the remaining figures from 9. Thus the arithmetical complement of log (55) or 1.740363 is 8.259637; and this number added to the logarithms of radius and of the side AC (43) will produce the same result as before, provided we reject ten from the integer part of the aggregate. Thus

log radius

log AC (43) 1.633469 arith. com. log AB (55) 8.259637

19.893106 The same result as before, if we reject 10 from the integer part of the sum.

Instead of rejecting ten from the result on account of the arithmetical complement, I recommend the pupil to incorporate the rejectaneous quantity by prefixing an unit to the arithmetical complement, and writing it in the form 18.259637. Here the figure i standing in the place of tens, is equivalent to ten, and the negative sign by which it is surmounted intimates that in the aggregation of the terms this single figure must be considered as negative, and consequently must be subtracted. The operation will stand thus

= 10.

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10.

1.633469 78.259637

9.893106 It is almost needless to add, that the arithmetical complement may be taken from the tables by inspection.

ÉXAMPLE 4. In a right-angled plane triangle, given the hypothenuse=100, and one of the acute angles=49°, required the base, perpendicular and remaining angle. Answer, Base 65.607, per

pendicular 75.47, and remaining angle 41'. EXAMPLE 5. Given the base of a right-angled plane triangle =70, and one of the acute angles=50°, required the perpendicular, hypothenuse and remaining angle. Answer, perpendicular

83.42, hypothenuse 108.90, and angle 40°. EXAMPLE 6. In a right-angled plane triangle given the hypothenuse 832 and base 768, required the perpendicular and acute angles. Answer, perpendicular 320, angles 67° 23', and 22° 37'.

With regard to oblique angled triangles, or those which have no right-angle, the solution of them may be reduced to the four following cases :

CASE I.

39. Given any two angles B, A, and a side BC, to find the other sides BA, AC. Solution. Construct the proportion

B

с BC sin B sin A : BC :: sin B : AC=

::sin C:AB, or::sin (A+B)

sin A : AB, (because sin C=sin (A + B).)

If BA had been the side given, we should have had sin (A+B) : BA:: sin B : AC :: sin A : BC.

Examples. In the oblique angled triangle ABC, given the angle A-889, the angle B=36°, and the side BC=56; required the other two sides.

BC x sin B From the above proportion we have AC=

sin A But log BC (56)

1.748188 log sin B (360) = 9.769219 arith. co. log sin A (889) = 10.000265 log AC

1.517672 whence AC=32.93

BC x sin (A+B Again AB=

sin A

log AB

But log BC (56)

1.748188 jog sin (A+B)=log sin (180°-A-B) =log sin 56°

9.918574 arith. co. log sin A

10.000265

1.667027 Whence AB=46.45.

EXAMPLE 2. Given the angle A=41° 13' 22", the angle C= 71° 19 5", and the side BC=55; required the other two sides.

Answer AB=79.063

AC=77.041 EXAMPLE 3. Given the angle B=78° 57', the angle C=47° 34', and the side BC=184; required the sides AC and AB.

Answer, AC=224.7, AB=169.

CASE II.

40. Given two sides, and an angle opposite to one of the given sides, to find the third side and the other two angles, supposing that we know of what species they are.

Solution. Suppose that the sides AB, AC are given, and the angle B; we shall first obtain the

B angle C by this proportion AC : sin B::AB : sin c, which will give the third angle. And we may then obtain the side BC by the proportion sin B: AC:: sin A : BC.

But to find the side BC at once, draw the perpendicular AD on the side BC; call AB, a; AC, b; sin B, s; cos B, c; and we shall have

B D
R: AB (a):: sin B (s): AD= : ; cos B (c): BD=

R

R Therefore DC=VAC?_AD=N 66

az

, and

R? act

? DC+BD=BC=R

bb_ EXAMPLE. In the oblique angled triangle ABC ( see the preceding figure) given AB=50, AC=40, and the angle B=32; required the side BC, and the angles A and C.

as

ac

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a?

2 S

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By the first Method. Sin C=

sin B x AB

AC
But log sin B (320) =

9.724210
log AB (50)

1.698970 arith. com. log AC (40) = 18.397940 log sin C

9.821120 Whence the angle C=41° 29', and angle A=106° 31'.

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