We shall illustrate this by a diagram, as it is a little, though only a very little, more complicated than the preceding ones. The only limitation required in this problem is, that the two given angles shall be together less than two right angles; for, if they are not, then they cannot be two angles of a triangle, as all the three are just equal to two right angles.

This being understood, let the line A в be the given side; and the angles at A and B, the given angles; it is required to construct the triangle.

First, let the two given angles be adjacent to the given side, A at the extremity A, and в at the extremity в ; and in this case it is not necessary to find the third angle, as it will be determined by the construction.

Make A B equal the given side; with any radius describe arcs on A and B, the given angles; and with the same radius describe arcs on the points A and B of the repeated line; cut off portions of these arcs, equal respectively to the arcs intercepted by the given angles A and B; and through a and B, the extre mities of a B, and the points of section, draw lines till they meet in c and c is the third angle, a c and в с the remaining sides, and the triangle is constructed.

B

The angles at A and B in this triangle are equal to the given angles at A and B, for by construction they stand upon equal arcs of equal circles. Also, the lines A c and в с must meet,

because the interior angles which they make with a в, toward the side c, are together less than two right angles. They must meet in some one point c, because two straight lines can meet in one point only; and as their positions are both determined by the given angles at A and B, and the point c must be in them both, they determine the place of the point c, and that point determines the length both of A c and в c; consequently the whole triangle is determined.

Secondly, Let one of the given angles-as, for instance, the angle c-be opposite the given side a B, and the other adjacent to it, as the angle a, at the extremity a of the given side; A B is the given side, and a and c the given angles.

The first thing to be done here, is to find the remaining angle at B; and after it is found, the triangle is constructed exactly as in the former case. To find the angle B,

draw any line D E, and in it take any point F, but not at either extremity. With any radius, describe arcs A or c on the given angles A and c; and with the same radius describe a semicircle on the line D E from F as a centre. Make the arc a of the semicircle equal to the arc of the given angle a, and the arc c of the same semicircle equal to that of the given angle c. Draw lines from F to mark the terminations of those arcs; and the remaining arc в represents the angle в of the triangle; for it is

the supplement of the given angles a and c to two right angles. Apply the arc a on the same radius to the extremity a of the given side, and the arc в to the extremity B of the same ; through A and B, and the extremities of the respective arcs, draw lines; and those lines will meet in c, and form the triangle required.

As the only difference between this case and the former is the finding of the angle в previous to the construction of the triangle, it requires no further explanation.

4. If two sides and an angle opposite one of them are given, the triangle may, in some cases, be constructed with the same certainty as by means of the data in any of the three preceding articles; but there are other cases in which the triangle obtained from these data is ambiguous, or admits of two different forms and magnitudes; and there are also cases in which the construction is impossible, so that there can be no triangle. The same ambiguity and the same impossibility will, of course, occur when we attempt to prove the equality of two triangles from two sides, and an angle opposite one of them; and therefore this mode of proof can be applied only in certain cases; and it is necessary to know what those cases are; for in them the proof is as clear and as complete as it is by any other data. We shall best explain this by taking data, and constructing the figures as we proceed.

A

A

A

A

Let a and c be the two given sides, and a the given angle. And first, let the angle a be adjacent to the less side a, and consequently opposite to the greater side c; and this is the determinate case in which the triangle can always be constructed, and has only one form and magnitude.

Draw any line A в, and produce it to such a length as may be judged convenient.

B

angle at a; and make a c equal in From c as a centre, and with a

Then, at the point a in the line a D, draw a c, making with a D an angle equal to the given length to the given side a. radius c в equal to the given side c, describe an arc cutting A D in the point в; join c B, and A B C is the triangle.

This may require some further explanation. The arc described with the greater of the given sides c as a radius, must cut the line a D if produced far enough; but it can do so only in one point between A and D; so that the point в cannot have two situations, and consequently the side a B, or the triangle A B C, cannot have two values.

To prove this: from c draw the dotted line c E perpendicular or at right angles to A B. CE is less than c B; because the angle CE B opposite C B is a right angle; and consequently the angle C B E must be less than a right angle, and the less side is opposite the less angle; consequently c E is less than C B ; and the point E must fall within the circle of which c is the centre, and CB the radius. But the point E is in the line a D, and therefore

the circle must cut or cross that line, before it can be on the opposite side of E from c.

Again: the point a must fall within the circle, because a c is by hypothesis, that is as the data were taken, less than c B the radius. Therefore the circle cannot cut A D a second time between A and D. · ́

The same figure will show us the state of the data which leads to the impossible case. If the given side c had been the less of the two, and less than the perpendicular c E let fall from c upon a D, a circle with this radius could not have cut the line A D; and consequently there could have been no point B, and no triangle.

Secondly, Let us examine the case in which the given angle is adjacent to the greater of the given sides, and opposite to the less. This is the ambiguous case in all states in which the opposite side is less than the adjacent one, but not less than the perpendicular; for when it is less than that, it becomes the impossible case.

A

A.

C

Let A and c be the two given sides, and a the angle adjacent to the side a, but let a be greater than c.