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10.

1.633469 18.259637

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 A 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. From the above proportion we have AC BC x sin B

=

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 sog sin (Ă +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.

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ac

:

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

A 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=as

: ; cos B (C): BD= R

R Therefore DC=VACP-AD2=

. and

R? act als DC+BD=BC=R'w bb=

R? Example. In the oblique angled triangle ABC (see the preceding figure) given AB=50, AC=40, and the angle B=320; required the side BC, and the angles A and C.

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=1060 31'.

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.

log BC

Sin AXAC Again, BC =

sin B But log sin A (106° 31')

9.981699 log AC (40)

1.602060 arith. com. log sin B (32) = 10.275790

1.859549 Whence BC=72.368

By the second Method we have BC=AB cos B+VAC-AB2 (sin B)?, and by substituting the values of AB, AC, sin B, cos B, we shall find BC=72.368, as before.

EXAMPLE 2. In the oblique angled triangle ABC let AB=120 feet, AC=42 feet, and the angle B=579 27; what are the other angles and the third side?

Answer, the angles are 64° 34' 23" and

51 58 58

and the side is 112. 653 feet. EXAMPLE 3. In the plane triangle ABC, let BC=345 feet AC=232 feet, and the angle B=37° 20', what are the other aro gles and the third side, supposing the triangle to be acute-angled, and what, if obtuse-angled? Answer, the angle A=64° 21' 1", or 115° 35' 59"

LC=78 15 59, or 27 41 1

and the side AB=374.559, or 174.073. N. B. The angle found by this method is ambiguous when the given angle is acute, and at the same time the side opposite the given angle is less than the other given side.

CASE III.

41. Given two sides and the included angle, to find the other two angles and the remaining side. Solution. Let the angle A and the two sides

A
AB, AC be given, and the angles B and C, and
the side BC be required.

B
First we have,
AC : AB :: sin B : sin C, or by composition
AC+AB : AC-AB ::sin B+sin C : sin E-sin C::

;1 sin B-sin

C sin B+sin C_tang ! (B+C)

sin B-sin Ctang $ (B-C) Therefore,

+

sin B+sin C:1

But (19)

AC + AB : AC—AB::tang (B+C) : tang (B-C)

A)= coton

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AC+AB:AC—AB :: cot1

And since B+C=180°-A, we have

1 tang (B+C)=tang (902 1C

A.

2 And therefore,

1
- cot. L, A : tang (B-C)

2 And consequently since we know B-C and B+C, it is easy to obtain the angles B and C; and we shall then have sin B: AC:: sin A : BC, which will give the third side BC.

This third case may also be solved in the following manner :
Draw the perpendicular BF upon the side
AC, and let AB=A, AC=0, sin A-s, cos
A=c, we shall have

'ac R:8:::

BF= ::C: AF-
R
R

B
Therefore FC-b But in tae right-angled triangle BFC,

R

А

F

as

ac

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Similarly tang B=a

Rabc

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ac

R2

R

R

1

2

If the angle A is obtuse, C becomes negative, and we have a Rs

bRs tangent C=

and tang B= 6 R tac

a R+bc The right-angled triangle BFC gives BC

a 52 =vden +(bm)? :} =v(a+b_2 abc); or w

R 2 (a* + 6* + 2 abc), if the angle A is obtuse

), 42. From the proportion

I AC+CB: AC—CB::tang (B+C): tang-(B-C) it follows that in any triangle the sum of any two sides is to their difference, as the tangent of half the sum of the two angles opposite these sides, is to the tangent of half the difference of these same angles.

Let then'AC=a, AB=d, (see preceding figure) the angle B=B, the angle C=C; we shall have

AR_P tang (B+C): tang}(B-C)::a+d: a-d::R: B+

a

9+ ÖR

R_R? :: R:

QR+R

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R:;R:d atd

a represents the tangent of any arc U, we shall

Therefore if

R

d
have
R tang U-RR

, or by (22)
tang
UHR

1

R:

R: tang (U—450): : tang (B+C): tang (B-C)

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we

tang Bag But AC +AB=882, AC=AB=242, and cor__n=tang B4C

Hence if we construct this proportion; the less side AB is to the greater AC :: R: tang arc U, and if we subtract 459 from this

1 angle, radius will be to tangent of the remainder, as tang=(B+C)

,

2 : tang (B-C).

2 Example 1. In the oblique angled triangle ABC (preceding figure) given AB=320, AC=562, and the angle A=1289 4; required the side BC.

By the first Method

cot A (AC-AB) have 2 AC+AB

+

2 180°_128° 4' = tang

:) =tang 25° 58'. 2

tang 25° 58' x 242 2

802 But log tang 25° 58' = 9.687540 log 242

2.383815 arith. co. log 882

17.054531
B.
log tang

9.125886

2 Whence the half difference of the angles B and C is 7° 36', and by a well known theorem 25058° +7° 36'=33° 34' will be the greater angle, and 25° 58-7° 36=18° 22=the less angle.

By the proportion sin B.: AC::sin A : BC, we shall then find the side BC=800.

As the second method is not adapted to logarithmic calculation, we shall not examplify it.

Therefore tang Ba

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By the Third Method we have

ACXR tang arc US

AB
But log AC (562)

2.749736
log radius
arith. co. log AB (320) =

17.494850 log tang arc U = 10.244586

= Whence arc U=600 20', and U-450=15° 20'

= 10.

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