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it will have two sides and an angle opposite one of them the same as those given. Hence, there will be two solutions if a has any value between the limits p and b.

2d. If a= = p, there will be but one solution.

3d. If a = b, there will be but one solution.

For, as a diminishes and approaches A p, the two points B and B' approach; and if a p, B and B' will unite, the arc will be tangent to c, and the two triangles will become one, and there will be one solution.

4th. If ab, there will be but one solution.

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a

5th. If a <p, there will be no solution.

b

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For, as a increases and approaches b, the points B and B' separate, the triangle ABC increases, and the triangle AB'C' decreases; and when a becomes equal to b, the triangle AB'C vanishes, and there remains but one triangle, or there is but one solution.

b p

A

a

a

a

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B

For, although there are two triangles ABC and AB'C, the latter is excluded by the condition that the given angle A is acute, since CAB' is obtuse, and there remains but one triangle ABC which satisfies the conditions, or there is but one solution.

B

For the arc described with C as

center and a as radius will neither intersect the opposite side nor be tangent to it. The triangle can not be constructed, or there will be no solution.

2. WHEN THE GIVEN ANGLE IS OBTUSE.

1st. If ab there will be but one solution.

с

For, although there are two triangles ABC and ABC, the latter is excluded by the conditions of the problem, since the angle CAB' is acute while the given angle is obtuse. There remains but one triangle, ABC, which satisfies all the conditions of the problem, or there is but one possible solution.

If ap the two triangles reduce to one, right-angled at B, which is excluded by the condition that the given angle is obtuse.

If a p no triangle can be constructed with the given parts, and there will be no solution.

2d. If a = b there will be no solution.

For as a diminishes and approaches b, B will ap proach A; and when a becomes equal to b, B will unite with A, and the triangle ABC will vanish. The triangle ABC will remain, but will be excluded by the conditions of the problem, since the angle CAB' is acute while the given angle is obtuse.

B

a

(

p a

a

p

3d. If a b there will be no solution; for then,

C

If a p there will be two triangles, AB'C and AB"C, but both are excluded by the condition that the given angle is obtuse.

B

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A

B

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a

A

A

B

A

72. Summary of Results.

1. When A 90°.

Two Solutions, If a>p and a <b.

1st. If a

One Solution,

{

2d. If a

No Solution, If a <p.

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= p.

b.

3d. If a b.
>

One Solution, If a b.

No Solution,

2. When A 90°.

b.

J 1st. If a =
2d. If a <b.

73. Method of Computation.

Reversing the order of the couplets of the proportion in Case I, we have

(1) ab sin A : sin B.

Hence, The side opposite the given angle is to the side opposite the required angle, as the sine of the given angle is to the sine of the required angle.

b sin A

a

(3) log sin Blog blog sin A+ a. c. log a - 10.

If there is but one solution, take from the table the angle B corresponding to log sin B; if there are two solutions, take B and its supplement B', for both correspond to log sin B.

We find C from the formula,

(1) gives (2) sin B=

C 180° (A + B) or C 180° (A+ B').
= -

We find c from the proportion,

sin A sin C:: a c,
:

a sin C

sin A

.. log clog a + log sin C+ a. c. log sin ▲ — 10.

1. Giv.

.'.

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1.09899

log b (12.56)
log sin A (30° 25′) = 9.70439
log P
0.80338

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pb sin A.

Introducing R and applying logarithms, we have

log plog blog sin A-10.

log b (12.56) 1.09899 log sin A (30° 25′)=9.70439 a. c. log a (9.25)

= 9.03386

log sin B

= 9.83724

p

b sin A

ab sin A: sin B, ... sin B a

S B
B'

p 6.3589.

Since ap and a <b, there are two solutions. Since the side opposite the given angle is to the side opposite the required angle as the sine of the given angle is to the sine of the required angle, we have the proportion,

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log sin Blog blog sin A+ a. c. log a 10.

a

B

43° 25′ 41′′.

136° 34' 19".

2. Given

C= 180° — (A + B) :

C' 180°

a sin C

sin A

log clog a log sin C+ a. c. log sin A-10.

Taking the value of C, we have,

log a (9.25)

log sin C (106° 9′ 19'')

sin A sin C: a c,

3. Given

4. Given

a. c. log sin 4 (30° 25')

log c

5. Given

a. c. log sin A (30° 25′)

log c

Taking the value of C', we have,

log a (9.25)

0.96614

log sin C' (13° 0′ 41′′)

9.35246

-0.29561

0.61421 ... c 4.1135.

a

b

A

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a

b

A

(A + B')

a

b

a =

b

20.35.

20.35.

52° 35′ 27′′.

645.8.

234.5.

48° 35'.

17.

40.25.

27° 43′ 15′′.

94.26.

=

}

= 126.72.

A=27° 50′.

106° 9′ 19′′,

13° 0′ 41′′.

}

.. C

0.96614

9.98250

0.29561

1.24425

B

Req. C

Req.

...C= = 17.549.

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52° 35′ 27′′.

74° 49′ 06′′.

24.725.

38° 52′ 16′′. 141° 7' 14".

CS 113° 17′ 14′′.

11° 2' 46".

=

No Solution.

S 185.439.

38.682.

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