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EXAMPLES. LXVII,

(1) If A=53° 24', B=66° 27', c=338.65 yards, find C and a, baving given that

L sin 53° 24' =9.9046168, log 3:3865 = -5297511,
L sin 60° 9'-9.9381851, log 3.1346=4961821,

log 3.1347= 4961960.

(2) If A=48°, B=54', and c=38 inches, find a and b, having given that

log 38=1.5797836, log 2.88704= -4604527, log 3.14295= .4973368, I sin 540=9.9079576,

L sin 78°=9.9904044, L sin 48°=9:8710735.

(3) Find c, having given that a=1000 yards, A=50°, C=66°, and that I sin 50°=9.8842540, L sin 66°=9.9607302,

log 1.19255=·0764762,

(4) Find b, having given that B=32° 15', C=21° 47' 20", a=34 feet.

log 3:4=531479, L sin 32° 15'=9.727228, log 2.241=•350442, L sin 54° 2' =9:908141, log 2.242=-350636, L sin 54° 3'=9 908233.

(5) Find a, b, c, having given A=72° 4', B=41° 56' 18", c=24 feet.

log 2-4

=:3802112, L sin 72° 4' =9:9783702, log 1•755 = •2442771, L sin 41° 56' 10"=9.8249725, log 1•756 = 2445245, L sin 41° 56' 20"=9.8249959, log 2:4995=-3978531, L sin 65° 59'

=9.9606739, log 2:4996='3978705, L sin 66° =9.9607302.

Case III.

258. Given two sides and the included angle, as b, c, A.

[Euc. I. 4; VI. 6.) First, B+C = 180o – A. Thus (B+C) is determined.

B-C

6-C A
Next,
tan

cot
2
b + c

2

Thus (B C) is determined.

And B and C can be found when the values of (B+C) and (B C) are known.

5

b.sin A
Lastly,
sin A

sin B

a

or a =

sin B'

Whence a is determined.

259. In practical work we proceed as follows;
B-C

A
Since tan

cot
2 b + c

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.: log (tan

dog (tan "79)+10

= log (6 –c) – log (6 + c) + log (cot 4) +
log (6 –c) – log (6+c)+L cot

+ 10,

B-C

or, L tan

.

2

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.. log a = log b + L sin A - L sin B, as in Case III.

Example. Given b=456.12 chains, c=296-86 chains, and A = 74° 20', find the other angles.

Here, b-c=159.26, 6+c=752.98.
From the Table we find,
log 159.26=2.2021067, and log 752.98=2.8767834,

L cot 37° 10'=10.1202593;

B-C
.: L tan =2.2021067 - 2:8767834+10:1202593

2

=9:4455826=L tan 15° 35' 18". ..B-C=31°10' 36", and B+C=180° — 74° 20'. Thus B+C=1050 40 ;

.: 2B=136° 50'36"; 2C=74° 29' 24", or,

B=68° 25' 18"; or, C=37° 14' 42". 260. The formula a* = + co – 2bc cos A may be used in simple cases; or it may be adapted to logarithmic calculation by the use of a subsidiary angle. [Cf. Example LXXVI. (20), p. 256.]

Example. If b=35 feet, c=21 feet, and A=50°, find a, given that cos 50o=.643.

Here a’=352 +212 – 2 x 35 x 21 x cos 50° ;

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a

=3.82 nearly; or, a=26•74=about 263 feet.

EXAMPLES. LXVIII.

(1) Find B and C, having given that A=400, b=131, c=72.

log 5.9 = 7708520, I cot 20° = 10:4389341,
log 2:03="3074960, L tan 38° 36'= 9:9021604,

L tan 380 37'= 9:9024195.

log 2

(2) Find A and B, having given that a=35 feet, b=21 feet, C=50°.

•301030, L tan 28° 11'=9.729020, L tan 650=10•331327, I tan 28° 12' =9.729323. (3) If b=19 chains, c=20 chains, A=60°, find B and C, having given that log 39= •591065, L tan 2° 32°=8·645853,

L cot 30°= 10·238561, L tan 2° 33'=8.648704.

(4) Given that a=376 375 chains, b=251•765 chains, and C=78° 26', find A and B. L cot 39° 13'=10.0882755,

log 1.2461=·0955529, L tan 13° 39'= 9:3853370,

log 6•2814=•7980565, L tan 13° 40'= 9.3858876. (5) If a=135, b=105, C=60°, find A, having given that

log 2=-3010300, L tan 12° 12' =9.3348711,

log 3=4771213, L tan 12° 13'=9.3354823. (6) If a=21 chains, b=20 chains, C=60°, find c. (7) Find c in the triangle of example (5).

(8) In a triangle the ratio of two sides is 5 : 3 and the included angle is 76° 30'. Find the other angles. log 2='3010300, L cot 35° 15' =10.1507464, L tan 19° 28' 50"

= 9:5486864.

Case IV.

261. Given two sides and the angle opposite one of them, as b, c, B. [Omitted in Euc. I; Euc. VI. 7.

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262. We remark however that the angle C, found from the trigonometrical equation sin C =a given quantity, where C is an angle of a triangle, has two values, one less than 90°, and one greater than 90°.

[Art. 234.] The question arises, Are both these values admissible? This may be decided as follows:

If B is not less than 90°, C must be less than 90°; and the smaller value for C only is admissible. If B is less than 90° we proceed thus.

csin B 1. If 6 is less than c sin B, then sin C, which

b is greater than 1. This is impossible. Therefore if 6 is less than c sin B, there is no solution whatever.

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2. If b is equal to c sin B, then sin C = 1, and therefore C = 90°; and there is only one value of C, viz. 90°. 3. If b is greater than c sin B, and less than

C,

then B is less than C, and C may be obtuse or acute. In this case C may have either of the values found from the equa

csin B tion sin C=

Hence there are two solutions, and the

6 triangle is said to be ambiguous.

4. If b is equal to or greater than c, then B is equal to or greater than C, so that C must be an acute angle; and the smaller value for C only is admissible.

.

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