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=

=

Sin AX AC 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 log BC

1.859549 Whence BC=72.368

By the second Method we have BC=AB cos B+VAC-AB’ (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=574 27; what are the other angles and the third side ?

Answer, the angles are 64° 34' 22" and

51 58 38

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 arr 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*

L C=78 15 59, or 27 411

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.

А A

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

I
First we have,
AC : AB :: sin B : sin C, or by composition

sin B+sin C 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 C tang (B-C) Therefore, AC + AB : AC-AB::tang

: tang

But (19)

:: cor

5

And since B+C=180°-A, we have

cot. A. 8 And therefore, AC+AB:AC-AB :: cot.

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=O, sin A--s, cos

F
A=c, we shall have
R:a::8:BF=

R
R

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

R we have

a R$
b

-
R
R

DR-ac

6 RS Similarly tang B=

a R-bc If the angle A is obtuse, C becomes negative, and we have a R.

6 Rs
and tang B=
6 R +ac

a R + bc The right-angled triangle BFC gives

as

'ac

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ac

tangent C=

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R2

2 abc (am + 6? + ), if the angle A is obtuse

R 42. From the proportion

I

1 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

a---d. (B+

R:;R:2 and

gt “R_R ' :: R:

R+R

[graphic]

Therefore if a represents the tangent of any arc U, we shall

bave
R:
R tang U-RR

-, or by (22)
tang U+R

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

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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).

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 1 A (AC-AB)
have
AC+AB

B+C

we

tang B But AC +AB=882, AC–AB=242, and cot_._ A=tang BAC.

180°—128° 4', =

° . B

tang 25° 58' x 242

802 But log tang 25° 58' 9.687540 log 242

2.383815 arith. co. log 882

– 17.054531 B.

-) = 9.125886

2 Whence the half difference of the angles B and C is 7° 36', and by a well known theorem 250:58° +7° 36'=33° 34 will be the greater angle, and 25° 58—7° 36=180 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.

log tang (BC

By the Third Method we have

tang arc U_ACx R

= 10.

AB
But log AC (562)

2.749736
log radias
arith. co. log AB (320) = 17.494850

log tang arc U = 10.244586 Whence arc U=60° 20', and U-450=15° 20'

[graphic]

B-C tang 15° 20' x tang 25° 58'
2

R
But log tang 25° 58' = 9.687540
log tang 15 20 9.438059

19.125599
log radius

10
B-C
log tang

9.125599

2 BC whence

70 36 2 The angles B and C, and the side BC will be found as before.

EXAMPLE 2. Given the two sides of an obtuse-angled plane triangle respectively 1280 and 1860, and the included angle=270 45'; required the remaining angles and side.

Ansner, the angles are 112° 54' 50",

and 39° 20' 10", and the side 940.237. EXAMPLE 3. In an acute-angled plane triangle, given the two sides AC, BC, respectively 120 and 112 yards, and the included angle C357° 58' 39"; what are the angles A, B, and side AB?

Answer, angle A=57° 27', B=640 34' 21",

and the side AB=112.653.

CASE IV.

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2 BC

2 a

43. Given the three sides of a triangle, required the three angles.

Let BC=0,...AB=b ... AC=d...R=1... and draw AD perpendicular on BC. By prop. 47, Book 1, Euclid's Elements

.B ABBDP-ACP_CD’=AC?—(BC-BD)?

AB? + BC-AC Whence BD

or BDC

a? +61Again in the right-angled triangle BDA, we have AB (6): BDA*+b?—–

-) :; rad (1): cos B-a'+b?—d? 2 a

2 ab d_o?+2

abB*_d?-(a+b)=2 sin f B(18) Therefore 1-cos B=

ab

2 ab From this expression we may deduce the following proportion: d-abd+a-6 x

:: R* : sinB; which, calling a the 2

2 semiper, meter, becomes ab : (9-a) (946) :: R’: sin’ { B; and by this expressinr we usually calculate the value of an angle from the three sides

We may also calculate it by means of the formula cos B R ab (ap+b?—d"), which gives

2 ab : a' +62_d?::R: cos B.

bB

ab :

2

[graphic]

B=2 ab

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That is, to find an angle B in a triangle, of which we know the three sides, we must first find the sum of the squares of the two sides containing the angle required, and subtract from this sum the square of the third side ; this will leave a remainder. We must then construct this proportion. The 'double of the rectangle of the side containing the angle required is to this remainder, as radius to the cosine of the angle required. If the angle B is obtuse, we shall have

R cos B

(d’_a_6%) 44. A formula equally convenient for logarithmic calculation, may be obtained as follows: We have just shewn that 1-cos B=2 sin?

.(9 2

ab a2 +62-od? (a+b)2--d2(a+b+d)(a+bood And 1 +cos B=1+

2 ab
2 ab

2 ab
B
ab

2 sin' B

(q-a) (9-)
2
B
ab

=(9a) (9–6, ; cog

2

۹ 9 (940); (9) (9-0)

ab
if radius=1 ; or
tang

B_R* (9-a) (9–6), if radius=R
2

9 (9-0) N. B. This formula must not be used, when the angle sought for is nearly 180°

-(4+b+ 2. 9 (9d)=2 cos?

by (18); consequently we have

.

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=

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Examples. 1. In a plane triangle given the three sides respectively 1756, 1214, 1992, required the angles. By the first formula we have

B _R? (9a) (9–6),
sin?
2

ab q representing the semiperimeter, and a=1756, b=1214, the siden about the angle B. 1

=2181 and qm=425, and qb=967

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Here q=* { 1756 +1214 +1392

2

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