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CHAPTER XVII.

ON THE SOLUTION OF TRIANGLES.

250. The problem known as the Solution of Triangles may be stated thus: When a sufficient number of the parts of a triangle are given, to find the magnitude of each of the

other parts.

251. When three parts of a Triangle (one of which must be a side) are given, the other parts can in general be determined.

There are four cases.
I. Given three sides.

[Compare Euc. I. 8.] II. Given one side and two angles. [Euc. I. 26.] III. Given two sides and the angle between them.

[Euc. I. 4.] IV. Given two sides and the angle opposite one of them.

[Compare Euc. VI. 7.]

Case I.

252. Given three sides, a, b, c. [Euc. I. 8; VI. 5.) We find two of the angles from the formulæ

А tan

b) (8 —c) 2

8 (8-a)

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log tan

=

log

8- a

253. In practical work we proceed as follows:

A (8 6) (8-c);
2

a) or,

A L tan 10 = 1 { log (8 6) + log (s – c) – log 8 - log (8-a)}.

2 Similarly,

B L tan - 10 =} {log (8 c) + log (8 a) – log 8– log (8 6)}.

А (8 b) (8 — c) 254. Either of the formulæ sin

2

bc

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A COS

2

8 (8

a)

may also be used as above.
bc
A

A
The sin

and the cos formulæ are either of them as 2

2

А convenient as the tan formula, when one of the angles only

2

, is to be found. If all the angles are to be found the tangent formula is convenient, because we can find the L tangents of two half angles from the same four logs, viz. log 8, log (8-a), log (8 b), log (8-c). To find the L sines of two half angles we require the six logarithms, viz. log (8 - a), log (8 - b), log (8 - c), log a, log b, log c.

Example. Given a=275-35, b=189-28, .c=301.47 chains, find A and B.

Here, 8=383.05, 8 - a=10770, 8- b=193•77, 8-c=81:58.
Then

A
L tan =10+}{log 193.77+log 81:58 - log 383.05 – log 107070}

2

=10+} {2.2872865+1:9115837 – 2:5832555 – 2:0322157} =9.7916995

[from the tables], А whence 31° 45' 28.5"; .. A=63° 30' 57".

Also

B L tan =10+} {log 81:58 + log 107.70 – log 383.05 – log 193.77}

2

=9.5366287=1 tan 18° 59' 9.8";
.: B=37° 58' 20"; C=1800 – A - B=78° 30' 43".

255. This Case may also be solved by the formula

72 + c -a

2bc

COS 4

But this formula is not adapted for logarithmic calculation, and therefore is seldom used in practice.

It may sometimes be used with advantage, when the given lengths of a, b, c each contain less than three digits.

Example. Find the greatest angle of the triangle whose sides are 13, 14, 15.

Let a=15, b=14, c=13. Then the greatest angle is A.

142 + 132 - 152 140 Now, cos A=

-=-384615.
2 x 14 x 13 2x 14 x 13
=cos 67° 23', nearly.

[By the Table of natural cosines.] .: the greatest angle=67° 23'.

EXAMPLES. LXVI.

(1) Ifa=352-25, b=513:27, c=482.68 yards, find the angle A, having given

log 674:10=2-8287243, log 321.85=2:5076535,

log 160·83=2.2063401, log 191-42=2.2819873,
L tan 20° 38' =9.5758104, L tan 20° 39'=9.5761934.

(2) Find the two largest angles of the triangle whose sides are 484, 376, 522 chains, having given that

log 6:91='8394780, log 3·15=4983106,

log 2:07="3159703, log 1.69=·2278867,
L tan 36° 46' 6"=9.8734581, I tan 31° 23' 9"=9.7853745.

(3) If a=5238, b=5662, c=9384 yards, find the angles A and B, having given

log 1.0142=·0061236, log 4.904= '6905505,

log 4:48=.6512780, log 7:58=:8796692,
L tan 14° 38'=9:4168099, I tan 15° 57'=9.4560641,
L tan 14° 39'=9:4173265, L tan 15° 58'=9:4565420.

(4) If a=4090, b=3850, c=3811 yards, find A, having given

log 5.8755=•7690448, log 3.85=-5854607,

log 1•7855= 2517599, log 3.811=:5810389, L cos 32° 15'=9.9272306, L cos 32° 16'=9.9271509. (5) Find the greatest angle in a triangle whose sides are 7 feet, 8 feet, and 9 feet, having given

log 3=·4771213, L cos 36° 42' =9.9040529, log 1'4=:146128, diff. for 60"= '0000942.

(6) Find the smallest angle of the triangle whose sides are 8 feet, 10 feet, and 12 feet, having given that log 2=-30103, L sin 20° 42' =9.5483585, diff. for 60" = .0003342. (7) If a :b:c=4:5:6, find C, having given

log 2=-3010300, log 3=4771213, L cos 41° 25'=9.8750142, diff. for 60" = .0001115. (8) The sides of a triangle are 2, 6, and 1+1/3, find the angles.

(9) The sides of a triangle are 2, 12 and 13-1, find the angles.

Case II.

256. Given one side and two angles, as a, B, C.

[Euc. I. 26; VI. 4.] First, A = 180° — B-C; which determines A,

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or,

.: log b = log a + log (sin B) + 10 – (10 + log sin A).

log b = log a + L sin B - L sin A. Similarly, log c = log a + L sin C - sin A.

Example. Given that c=1764:3 feet, C=18° 27', and B=66° 39', find b. From the Tables we find log 1764:3=3.2465724. L sin 18° 27' =9.5003421, L sin 66° 39'=9.9628904 ; .: log b=3.2465724+9.9628904 – 9.5003421

=3•7091207=log 5118.2 ; ..b=5118.2 feet.

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