This section is from the book "The American House Carpenter", by R. G. Hatfield. Also available from Amazon: The American House Carpenter.

To find the two sides: if a be the given side, then to find the side b we have, equation (193.) -

b = a sin. B/sin. A;

or, the side b equals the product of the side a into the quotient obtained by a division of the sine of the angle opposite b by the sine of the angle opposite a.

For example: in a triangle (Fig. 318) in which the angle A = 18°, the angle B = 42o (and, consequently (Art. 345) the angle C= 120o), and the given side a equals 43 feet; what are the lengths of the sides b and c? Equation (193.) gives -

b = a sin. B/sin. A

Performing the problem by logarithms (Art. 427), we have -

Log. a(=43) | = 2.6334685 |

Sin. B (= 420) | = 9.8255109 |

1.4589794 | |

Sin. A (= 18°) | = 9.4899824 |

Log. b(=93-1102) | = 1.9689970. |

Thus the side 6 equals 93.1102 feet, or 93 feet 1 inch and nearly one third of an inch.

For the side c, we have, equation (195.) -

c = a sin. C

sin. A

or -

Log. a (= 43) | = 1.6334685 |

Sin. C(= 120o) | = 9.9375306 |

1.5709991 | |

Sin. A (= 18o) | = 9.4899824 |

Log. c(= 120.508) | = 2.0810167 |

or, the base c equals 120 feet 6 inches and one tenth of an inch, nearly. But if instead of a the side b be given, then for a use equation (191.), and for c use equation (196.).

And, lastly, if c be the given side, then for a use equation (192.), and for b use equation (194.).

478. - Oblique-Angled Triangles: Second Class. - The problems which comprise the second class are those in which two sides and an angle opposite to one of them are given, to find the two remaining angles and the third side.

The only requirement really needed here is to find a second angle; for, with this second angle found, the problem is reduced to one of the first class; and the third side may then be found under rules given in Art. 477.

To find a second angle, use one of the equations (197.) to (202.).

For example: in the triangle ABC (Fig. 318), let a(= 43) and b(= 93.11) be the two given sides, and A, the angle opposite a, be the given angle (= 18o). Then to find the angle B, we have equation (199.) - (selecting that which in the right hand member contains the given angle and sides) -

sin. B = sin. A b/a

= sin. A 93.11/43.

By logarithms (Art. 427), we have -

Log. | sin. A(= 18o) | = 9.4899824 |

,, | 93.11 | = 1.9689970 |

,, | 1.4589794 | |

,, | 43 | = 1.6334685 |

,, | sin. B (= 420) | = 9.8255109 |

By reference to the log. tables, the last line of figures, as above, is found to be the sine of 42o; therefore, the required angle B is 42o. Then 180o - (18o + 42°) = 120o = the angle C.

With these angles, or with any two of them, the third side c may be found by rules given in Art. 477.

Fig. 319.

479.__Oblique-Angled Triangles: Sum and Difference of Two Angles. - Preliminary to a consideration of problems in the third class of triangles, it is requisite to show the relation between the sum and difference of two angles.

In Fig. 319, let the angle A JM and the angle A JN be the two given angles; and let A JM be called angle A, and A JN, angle B. Now the sum and difference of the angles may be ascertained by the use of the sum and difference of the sines of the angles, and by the sum and difference of the tangents. In the diagram, in which the radius A J equals unity, we have MP, the sine of angle A (= A J M), and NQ = RP, the sine of angle B(= A J N). Then -

MP- RP=MR

equals the difference of the sines of the angles; and since PM' = PM -

PM'+RP= RM',

equals the sum of the sines of the angles.

With the radius JC describe the arc JDE, and tangent to this arc draw FH parallel with MM', or perpendicular to A B.

Then FD is the tangent of the angle MCN, and D H is the tangent of the angle NCM'.

Now since an angle at the circumference is equal to half the angle at the centre standing on the same arc (Art. 355), therefore the measure of the angle MCN is the half of MN, equals -

1/2(AM - AN)=1/2(A-B).

Similarly, we have -

1/2 (AM'+AN) = 1/2(A +B),

for the angle NCM'.

Therefore we have for the tangent of the angle MCN -

FD = tan. 1/2(A -B),

and, for the tangent of the angle NCM' -

DH= tan. 1/2(A + B).

And, because FCD and MCR are homologous triangles, as, also, D CH and R'CM', therefore -

M'R: MR:: DH: DF,

sin. A + sin. B: sin. A - sin. B:: tan. 1/2(A + B): tan. 1/2(A - B),

from which we have -

sin. A - sin. B/sin. A + sin. B = tan. 1/2 (A - B)/tan. 1/2 (A + B.)

To obtain a proper substitute for the first member of this expression we have, equation (195.) -

c = a sin. C/sin. A or -

c sin. A = a sin. C. (M.)

We also have, equation (196.) -

c = b sin. C/sin. B' or -

c sin. B = b sin. C. (N.)

These two equations, (M.) and (N.), added, give -

c sin. A + c sin. B = a sin. C + b sin. C. or -

c (sin. A + sin. B) = sin. C(a + b). (P.)

But, if equation (N.) be subtracted from equation (M.), we have -

c sin. A - c sin. B = a sin. C - b sin. C,

or -

c(sin. A - sin. B) = A sin. C(a - b). (R.)

If equation (R.) be divided by equation (P.), we have -

c(sin. A - sin. B)/c(sin. A + sin. B) = sin. C(a - b)/sin. C(a + b),

which reduces to -

sin. A - sin./sin. A + sin. B=a-b/ a + b'

The first member of this equation is identical with the first member of the above equation (D.), and therefore its equal, the second member, may be substituted for it; thus -

a-b/a+b = tan. 1/2 (A-B)/tan. 1/2 (a+B),

From which we have -

tan. 1/2 (A-B) = tan. 1/2(A+B) a-b/a+b. (203.)

We have (Art. 431) the proposition, that if half the difference of two quantities be subtracted from half their sum, the remainder will equal the smaller quantity. For example: if A represent the larger quantity and B the smaller, then -

1/2(A+ B) -1/2(A -B) = B; (204.)

and, again, we also have (Art. 431) -

1/2(A+B) + 1/2(A-B)=A. (205.)

480. - Oblique-Angled Triangles: Third Class. - The third class of problems comprises all those cases in which two sides of a triangle and their included angle are given, to find the other side and angles.

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