Ratio of the sum of the two sides to their difference.

5. Given two sides of a triangle equal to 77.245 and 92.341, and the angle opposite the first side equal to 124° 31′48′′; to solve the triangle.

Ans. The question is impossible.

6. Given two sides of a triangle equal to 75.486 and 92.341, and the angle opposite the first side equal to 55° 28′ 12′′; to solve the triangle.

Ans. The question is impossible.

81. Theorem. The sum of two sides of a triangle is to their difference, as the tangent of half the sum of the opposite angles is to the tangent of half their difference. [B. p. 12.]

Proof. We have (fig. 1.)

a: b = sin. A: sin. B;

whence, by the theory of proportions,

a+b: a b sin. Asin. B : sin. A

=

which, expressed fractionally, is

Given two sides and the included angle.

or

a+b: a b tang. † (A+B) : tang. † (A — B).

82. Problem. To solve a triangle when two of its sides and the included angle are given. [B. p. 43.]

Solution. Let the two sides a and b (fig. 1) be given, and the included angle C; to solve the triangle.

First. To find the other two angles. Subtract the given angle C from 180°, and the remainder is the sum of A and B, for the sum of the three angles of a triangle is 180°; that

is,

and

A+B 180° - C,

=

=

(A+B) = 90° - C complement of C.

2

The difference of A and B is then found by (136)

1

a+b: a b tang. } (A+B): tang. ¦ (A— B). But we have

The greater angle, which must be opposite the greater side, is then found by adding their half sum to their half difference; and the smaller angle by subtracting the half difference from the half sum.

Secondly. The third side is found by the proportion

sin. A siu. C= a: c;

whence

Given two sides and the included angle.

a sin. C
sin. A

83. EXAMPLES.

1. Given two sides of a triangle equal to 99.341 and 1.234, and their included angle equal to 169° 58′; to solve the triangle.

Solution. Making a = 99.341, b = 1.234; and

Segments of base made by perpendicular from opposite vertex.

2. Given two sides of a triangle equal to 0.121 and 5.421 and the included angle equal to 1°2′; to solve the triangle. Ans. The other side = 5.3

84. Theorem. One side of a triangle is to the sum of the other two, as their difference is to the difference of the segments of the first side made by a perpendicular from the opposite vertex, if the perpendicular fall within the triangle; or to the sum of the distances of the extremities of the base from the foot of the perpendicular, if it fall without the triangle. [B. p. 14.]

Proof. Let AC (figs. 12 and 13) be the side of triangle ABC on which the perpendicular is let fall, and BP the perpendicular.

From B as a centre with a radius equal to BC, the shorter of the other two sides, describe the circumference CC EE. Produce AB to E' and AC to C', if necessary.

Then, since AC and AB are secants, we have,

(fig. 13.) AC = AP + PC = AP + PC

Given the three sides.

whence

(fig. 12.) AC: AB + BC = AB

=

(fig. 13.) AC: AB + BC AB BC: AP + PC.

85. Problem. To solve a triangle when its three sides are given. [B. p. 43.]

Solution. On the side b (figs. 2 and 3) let fall the perpendicular BP.

(fig. 3.) b caca: PA+ PC.

difference of the segments Then, adding the half dif

These proportions give the (fig. 2), or their sum (fig. 3). ference to the half sum, we obtain the larger segment corresponding to the larger of the two sides a and c. And, subtracting the half difference from the half sum, we obtain the smaller segment.

Then, in triangles BCP and ABP, we have, by (4) and

The third angle B is found by subtracting the sum of A

and C from 180°.