Elements of Plane and Spherical Trigonometry

from an assumed point a in which draw a line to make the angle A =22° 37'. Make AC = 216, and from c as a centre with a radius =117 describe an arc BB', it will

Β'

cut the line AB in two points, from each of which drawing lines BC, B'C, there will be formed two triangles ABC, ABC, each of which answer the conditions of the question. The required lines and angles being measured, give

....

AB = 117.... ACB = 2210 ABC 13430
AB'= 282.... ACB' 112°
=

....

Computation. To find the angle в.

AB'C =

4510.

As BC.... 117...... 2.0681859
To sin A 22° 37'....

So is AC ..

9.5849685

216...... 2:3344538

To sin в ...... 45° 14' or 134° 46′.. 9·8512364

Add to each A 22° 37′.. 22° 37′

Remain ... 112° 9′ or 22° 37′, being ACB' and ACB. Here since ACB BAC, we have AB = BC = 117. But to find AB, reverse the first two terms of the former proportion, and say,

As sin A

22° 37′.... 9.5849685

To BC.. ... . . 117 . . . . . . . .
So is sin ACB'..112° 9′

....

2.0681859

9.9667048

TO AB'......281·79...... 2·4499222

6. Remark. The ambiguity in this and similar examples, does not, as has been often affirmed, depend upon the circumstance that an angle and its supplement have the same sine, but solely upon this, that in con

structing triangles from analogous data, when the side CB has its length between certain limits, that is, between the length of the side AC, and that of the perpendicular cd from c on the third side AB, it must necessarily cut the indefinite right line ABB' in two points. Beyond those limits there is no ambiguity; for when CB is proposed to be less than cd, the problem is impossible; and when CB exceeds CA, the angle A being all along supposed given, the line ABB' can only be cut in one point. Hence the practical maxim may be thus expressed :—when cв is proposed less than AC sin A, the data are erroneous; when it is given greater than AC there can be only one triangle; between those limits the problem is ambiguous.

Ex. 3. Given two angles of a plane triangle 22° 37' and 134° 46′, and the side between them 351. To find the remaining angle and sides.

Ans. Angle 22° 37′; sides 351 and 648.

Ex. 4. Given two sides of a plane triangle 50 and 40 respectively, and the angle opposite to the latter equal to 32°; to determine the triangle.

Ans. If the angle opposite to the side 50 be acute, then is it = 41° 28', the third angle 106° 32′, and the remaining side 72:36. If the angle opposite to side 50 be obtuse, it is = 138° 32′, and the other angle and side 9° 28′ and 12-415 respectively.

CASE II.

7. When two sides and the included angle are given. The solution is effected by means of chap. ii. prop. 15. Take the given angle from 180°, the remainder will be the sum of the other two angles,

Then say,-As the sum of the given sides,

Is to their difference;

So is the tangent of half the sum of

the remaining angles,

To the tangent of half their difference.

Half the difference added to half the sum of those angles, gives the greater of them; and taken from half the sum, leaves the less *.

All the angles becoming known by this process, the third side is found by the rule in case 1.

Example I.

8. Let there be given in a plane triangle ABC, AC 450, BC= 540, and the included angle c = 80°; to find the remaining angles and side.

Leaving the construction to be effected by the pupil, I shall proceed to the computation.

BC AC 990, BC - AC 90, 180° — c = 100°. ...990.. 2-9956352 90.. 1.9542425

AS BC + AC

To BC

So is tan (A+B) 50°.. 10 0761865

To tan (AB)..6° 11′ 90347938

Hence 50°+6° 11′ 56°11′A; 50°-6°11'=43°49'.
Then, to find the third side AB, say,
As sin B.... 43° 49′..9.8403276
To AC ......450....2.6532125
So is sin c..80° 0′..9·9933515

To AB...... 640 08.. 2·8062364

Example II.

9. The two sides of a triangle are 40 and 32, and the included angle 90°, required the other angles and side. In this example the operation may be considerably shortened by working without, instead of with, the loga

=s+d,

* Let a + b =s, and a − b = d: then, by addition 2a: or a = as+d; and, by subtraction 2b = sd, or b = 4s — 1d.

rithms. For, since the given angle is 90°, the half sum of the remaining angles is 45°, whose natural tangent is unity; and the sum and the difference of the given sides are 72 and 8 respectively. Hence

As 72:8::1: =

=·1111111=nat. tan. 6° 20′.

Therefore 45° + 6° 20′ 51° 20, and 45° 6° 20′ 38° 40′, are the remaining angles.

And the third side (Euc. i. 47) = √ 402 + 322 √2624 = 8 √/41 = 51·225.

**Other methods, still shorter, of solving right angled triangles, will be given before we terminate the present chapter.

Ex. 3. Given two sides of a plane triangle 1686, and 960, and their included angle 128° 4′; to find the rest. Ans. Angles 33° 35', 18° 21', side 2400.

CASE III.

10. When the three sides of a plane triangle are given, to find the angles.

1st Method. Assume the longest of the three sides as base, then say, conformably with chap. ii. prop. 16. As the base,

To the sum of the two other sides;

So is the difference of those sides,

To the difference of the segments of the base.

Half the base added to the said difference, gives the greater segment, and made less by it gives the less; and thus, by means of the perpendicular from the vertical angle, divides the original triangle into two, each of which falls under the first case.

2d Method. Find any one of the angles by means of prop. 17, of the preceding chapter; and the remaining angles either by a repetition of the same rule, or by the relation of sides to the sines of their opposite angles.

Example I.

11. The three sides of a triangle are 40, 34, and 25.