Sidebilder
PDF
ePub

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.

S

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

A (8-6) (8-0)
tan
2

8 (8-a)
B
tan

(8 – c) (8 – a).

c 2

8 (8-6) The third angle C = 180 – A B.

[ocr errors]

b) (8–c);

log tan

253. In practical work we proceed as follows:

A
log

(
2

8 (8-a) or,

A L tan

- 10=1 {log (8-6) + log (8 -c) - log 8 - log (8 - a)}. 2 Similarly,

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

. 2

А

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

2

bc

an

8

[ocr errors]
[ocr errors]

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 s, log (8-a), log (8–6), log (8 —c). To find the L sines of two half angles we require the six logarithms, viz. log (8 - a), log (8 – 6), 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, s= =383.05, 8 - a=107.70, -b=193.77, 8-c=81:58.
Then

A
L tan =10+} {log 193.77+log 81:58 - log 383.05 – log 10770}

2

=10+} {2-2872865 +1.9115837 – 2.5832555 – 2:0322157} =97916995

[from the tables],

whence

4–3

= 31° 45' 28.5"; .. A=63° 30' 57".

Also

[ocr errors]

=

B L tan

2

10+} {log 81.58 +log 107-70 – log 383.05 – log 193.77}

=9.5366287=L tan 18° 59' 9.8";
.. B=37° 58' 20"; C=180° – A – B=78° 30'43".

[ocr errors]

255. This Case may also be solved by the formula

72 + c -a

2bc 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 =

t='384615.
2 x 14 x 13 2 x 14 x 13
=cos 67° 23', nearly.

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

[ocr errors]

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,
I 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 315=-4983106,

log 2:07="3159703, log 1.69=·2278867,
L tan 36° 46' 6"=9.8734581, L 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, L tan 15° 57'=9:4560641,
L tan 14° 39'=9-4173265, L tan 15° 58'=9.4565420.

a=

=

(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) angles.

The sides of a triangle are 2, 76, and 1+1/3, find the

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

Case II.

[ocr errors]

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

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

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small]
[blocks in formation]

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

log b = log a + L sin B I sin A. Similarly, log c = log a + L sin C - L 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 51182; .:b=5118.2 feet.

« ForrigeFortsett »