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Limitation. The two given angles must be together less than two right angles, or the lines PQ, QR would not meet. This follows also from the theorem that any two interior angles of a triangle are together less than two right angles, and is the contrapositive of that theorem.

PROBLEM 9.

To construct a triangle, having given two angles and a side opposite to one of them.

Let A and B be the given angles, CD the given side which is to be opposite to A.

Construction. Draw an indefinite straight line EF At any point G in it make the angles FGH = A, and HGK=B (Prob. 6), then KGE will equal the third angle of the triangle, since the sum of the three angles of

K

B

a triangle is equal to two right angles (Th. 25). At C and D make angles equal to HGK and KGE, and let their sides meet in O; then OCD is the triangle required.

Proof. For OCD has CD equal to the given line, and the angles C and D equal respectively to the given angles.

Limitation.-As before, the two given angles must be together less than two right angles.

PROBLEM IO.

To construct a triangle, having given two sides and the angle between them.

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Let A, B be the given sides, C the given angle.

Construction. Draw an angle D equal to the given angle, and take DE, DF equal to A and B. Join EF.

Proof. For the triangle DEF has DE, DF equal to the given lines A and B, and the included angle D equal to the given angle C.

Remark. In these problems we have found that one triangle and only one can be constructed to fulfil the conditions given. In other words, that with these data the triangle is determinate. Also we notice that in each case three elements in the triangle are data or given. We have given either the three sides, or two angles and the side adjacent to both, or two angles and a side opposite to one, or two sides and the included angle. And these cases correspond to the theorems proved above of the equality of triangles. For if only one triangle can be constructed so as to have its sides equal to three given lines, it is clear that if two triangles have the three sides of the one equal to the three sides of the other, these triangles must be identical, or be equal in all respects. And a similar remark may be made on the other cases we have considered.

But there are cases in which the data may be insufficient to determine the triangle. For example, if only two sides are given, an

indefinite number of different triangles may be constructed to have these sides. Or if the three angles are given, their sum being equal to two right angles, an indefinite number of triangles may be constructed to have these three angles. And again it may be impossible to construct the triangle with the given data, as has been already shewn. In some cases moreover the solution is ambiguous, that is, there may be more than one triangle which fulfils the given conditions. The following is an important instance of this, and is usually called the ambiguous case, some consideration of which occurred in Theorem 20.

PROBLEM II.

To construct a triangle, having given two sides and an angle opposite to one of them.

Let A, B be the given sides, C the angle to be opposite

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Take an angle GDHC, take DE = A, and with centre E and radius = B describe a circle. If I is one of the points in which this circle meets the line DH, by joining EI we obtain a triangle which fulfils the given conditions.

But several cases may arise.

Let the given angle be acute, as in the figure.

Then, by Theorem 19,

(1) If B is less than the perpendicular from E on DH, the circle would not meet DH, and the triangle would be impossible.

(2) If B is equal to the perpendicular, the circle would meet DH at the foot of the perpendicular, and there would be one triangle, right-angled, which fulfils the given conditions.

(3) If B is greater than the perpendicular but less than DE, then the circle will meet DH in two points I, I' as in the figure, on the same side of D, and there will be two triangles EDI, EDI' which fulfil the given conditions.

(4) If B is equal to DE, the point I will coincide with D, and one of the two triangles disappears, and the other is isosceles.

(5) If B is greater than DE, the circle will meet DH in two points on the opposite sides of D, but one only of the triangles made by joining EI, EI' will be found to have the angle D, and the other will have the supplementary angle: that is, there will be only one solution.

The cases of the given angle being a right angle or an obtuse angle are left to the ingenuity of the student.

SECTION V.

Loci.

WHEN a point has to be found to fulfil one given geometrical condition, the problem is indeterminate: that is, an infinite number of points can be found to fulfil the given condition. For example, if the problem is to find a point at a given distance from a given point, it is plain that all the points in the circumference of a circle, described with that point as centre and the given distance as radius, fulfil this condition. Or again, if a point has to be found at a given distance from a given straight line of indefinite length, it may lie anywhere on either of two straight lines parallel to the given line, and at the given distance from it on either side.

All the points which satisfy a single given geometrical condition lie in general in a line or lines and this line, or these lines, are called the locus of the point under the given condition. Hence we get the following definition of a locus :

Def. If any and every point on a line or group of lines (straight or curved), and no other point, satisfies an assigned condition, that line or group of lines is called the locus of the point satisfying that condition.

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