3. In the triangle ABC, are given the angle C = 33° 21', the side BC = 95.12 and the side AB = 60, to find the angles A and B, and the side AC.

By Construction, Fig. 44.

This is constructed in the same manner as the preceding example; only AB, being shorter than BC, the arc ab cuts AC in two points on the same side of BC; hence the angle A may be either acute or obtuse. The side required has also two values as AC and AC.

The sum of the angles C and A subtracted from 180° leaves the angle B = 86° 1' if A be acute, or 27° 17′ if A be obtuse.

To find the side AC answering to the acute value of the angle A.

To find the side AC, answering to the obtuse value of the angle A.

4. In a triangle ABC, the side AB is 274, AC 306, and the angle B 78° 13'; required the angles A and C, and the side BC. Ans. A = 40° 33', C = 61° 14', and BC203.2.

=

5. In a right angled triangle, there are given the hypothenuse AC 272, and the base AB=232; to find the angles A and C, and the perpendicular BC. Ans. A = 31° 28′, C= 58° 32′ and BC = 142.

6. In a right angled triangle ABC, the hypothenuse AC is 150 and one side BC 69; required the angles and other side. Ans. C = 62° 37′, A = 27° 23′ and AB 133.2.

CASE 3.

Two sides and the included angle being given, to find the other angles and side.

RULE.

Subtract the given angle from 180°, and the remain

As the sum of the two given sides,

Is to their difference;

So is the tangent of half the sum of the two unknown angles,

To the tangent of half their difference.*

This half difference of the two unknown angles, added to their half sum, will give the angle opposite the greater

* DEMONSTRATION. Let ABC, Fig. 45, be the proposed triangle, having the two given sides AB, AC, including the given angle A. About A as a centre, with AC the greater of the given sides,, for a distance, describe a circle meeting AB produced in E and F, and BC in D; join DA, EC, and FC, and draw FG parellel to BC, meeting EC produced, in G.

The angle EAC (32.1.) is equal to the sum of the unknown angles ABC, ACB, and the angle EFC at the circumference is equal to the half of EAC at the centre (20.3.); therefore EFC is half the sum of the unknown angles; but (32.1.) the angle ABC is equal to the sum of the angles BAD and ADB or BAD and ACB; therefore FAD is the difference of the unknown angles ABC, ACB, and FCD, at the circumference is the half of that difference; but because of the parellels DC, FG, the angles GFC, FCD are equal; therefore GFC is equal to half the difference of the unknown angles ABC, ACB; but since the angle ECF in a semicircle, is a right angle, EG is perpendicular to CF, and therefore CF being radius, EC, CG are the tangents of the angles EFC, CFG; it is also evident that EB is the sum of the sides BA, AC, and that BF is the difference; therefore since BC, FG are parellel EB: BF::EC: CG (2.6.): that is, the sum of the sides AC, AB, is to their difference, as the tangent of half the sum of the angles ABC, ACB, is to the tangent of half their difference.

To demonstrate the latter part of the rule, let AC and AB, Fig. 46, represent any two magnitudes whatever; in AB produced, take BD equal to AC the less, and bisect AD in E.

Then because AE is equal to ED and AC to BD, CE is equal to EB; therefore AE or ED is half the sum of the given magnitudes AB, AC, and CE or EB is half their difference; but AB the greater is equal to AE, EB, that is to half the sum added to half the difference, and AC the less, is equal to the excess of AE, half the sum, above

of the two given sides, and being subtracted from the half sum, will give the angle opposite the less given side.

After finding the angles, the other side may be found by Case 1.

EXAMPLES.

1. In the triangle ABC, there are given AB = 128, AC 90, and the angle A= 48° 12', to find the angles B and C, and the side BC.

=

=

By Construction, Fig. 47.

Draw AB 128, and make the angle A= 48° 12'; draw AC 90, and join BC. The angle B will measure 44° 37', the angle C 87° 11', and the side BC 95.5.

=

2.33846

- 1.57978

As the sum of the sides AB, AC, 218

Is to their difference, 38

So is the tang. of half sum of angles B & C, 65° 54' 10.34938

To tang. of half their difference, 21° 17′

11.92916

- 9.59070

65° 54'

21 17

87 11

Extend the compasses from 218, the sum of the sides, to 38, their difference, on the line of numbers, and apply this extent to the line of tangents from 45° to the left hand; then keeping the left leg of the compasses fixed, move the other leg to 65° 54′ the half sum of the angles; that distance will reach from 45° on the same line, to 21° 17', the half difference of the required angles. Whence the angles are obtained as before.

To extend the second proportion, proceed as directed in Case 1st.

2. In a triangle ABC, are given AB = 109, BC = 76, and the contained angle B = 101° 30′, to find the other angles and side. Ans. The angle A = 30° 57′, C = 47° 33', and the side AC = 144.8.

3. Given, in a right angled triangle, the base AB: 890 and the perpendicular BC = 787, to find the angles and hypothenuse. Ans. The angle A = 41° 29′ C = 48° 31′ and the hypothenuse AC = 1188.