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IX. SOLUTION OF OBLIQUE TRIANGLES.

120. In the solution of plane oblique triangles we may distinguish four cases:

1. Given a side and any two angles.

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2. Given two sides and their included angle.

3. Given the three sides.

4. Given two sides and the angle opposite to one of them. S

CASE I.

121. Given a side and any two angles.

The third angle may be found by Geometry, and then by aid of Art. 113 the remaining sides may be calculated.

The triangle is always possible for any values of the given elements, provided the sum of the given angles is < 180°.

1. Given b = 20, A = 104°, B = 19°.

Find a, c, and C.

(A+B) = 180° - 123° 57°.

=

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That is,

a=b sin A csc B, and c =

b sin C csc B.

Whence, log a = log b + log sin A + log csc B,

log clog blog sin C+ log csc B.

and

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Note. To find the log cosecant of an angle, subtract the log sine from 10 - 10. To find log sin 104°, take either log cos 14° or log sin 76°. (See pages 7 and 10 of the explanation of the tables.)

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122. Given two sides and their included angle.

Since one angle is known, the sum of the remaining angles may be found, and then their difference may be calculated by aid of Art. 114.

Knowing the sum and difference of the angles, the angles themselves may be obtained, and then the remaining side may be computed as in Case I.

The triangle is possible for any values of the data.

6

130°

1. Given a = 167, c=82, B = 98°. Find A, C, and b.

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Whence, log tan (4- C)= log (ac)+colog (a+e)

+log tan (44 C).

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Hence,

and

.. (AC) = 16° 31.7'.

A = {(A + C) + } (A − C ) = 57° 31.7',
C=(4+ C)-(4-C)=24° 28.3'.

To find the remaining side, we have by Art 113,

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CASE III.

123. Given the three sides.

The angles might be calculated by the formulæ of Art. 117; but as these are not adapted to logarithmic computation, it is more convenient to use the formulæ of Art. 118.

Each of the three angles should be computed trigonometrically, as we then have a check on the work, since their sum should be 180°.

If all the angles are to be computed, the tangent formulæ are the most convenient, as only four different logarithms are required. If but one angle is required, the cosine formulæ will be found to involve the least work.

The triangle is possible for any values of the data, provided no side is greater than the sum of the other two.

If all the angles are required, and the tangent formulæ are used, they may be conveniently modified as follows:

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1. Given a 2.5, b = 2.79, c = 2.33; find A, B, and C.

In this case,

Whence,

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By logarithms, we have

log r=[log (sa) + log (sb) + log (sc)+colog s].

Also,

log tan 4= log r- log (sa),

log tan Blog r — log (s — b),

log tan Clog r — log (s— c).

log (sa)=0.1173

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logr = 9.8576-10

log (sb)=0.0086

log tan B 9.8490 - 10

=

B=35° 14.1'

.. B=70° 28.2'.

log r9.8576 - 10 log (sc) = 0.1703

log tanC 9.6873-10

Check, A+B+C=179° 59.2'.

=

C=25° 57.2'

...C=51° 54.4'.

2. Given a = 7, b = 11, c = = 9.6; find B.

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Or, log cos B = [log s + log (s—b)+colog c+colog a].

In this case,

Whence,

2s=a+b+c=27.6.

s = 13.8, and s-b= 2.8.

log s = 1.1399

log (s—b) = 0.4472

colog c=9.0177 – 10

colog a = 9.1549 — 10

2)19.7597-20

log cos B 9.8798 - 10

B = 40° 41.8'

... B=81° 23.6'.

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