W. L. Nekrasov. The Construction of Triangles on the Sphere.

V. L. Nekrasov

Construction

of Triangles

on the Sphere.

Tomsk.

Typo-Lithography of the Siberian Society of Printing Business, corner of Dvoryanskaya St. and Yamskoy Lane, own building.

1911.

CONSTRUCTING TRIANGLES ON A SPHERE.

The literature on spherical geometry and spherical trigonometry is immense, and at the present time it is difficult to say anything that has not already been said at some time and by someone before. Nevertheless, my searches in this respect have not yielded the desired result: what I set forth in the present article I have not found in other authors; true, although I have looked through much, I have not seen the greater part of the literature; the reason for this is that obtaining the relevant articles and works, often published very long ago and in little-circulated journals, is far from easy; they are absent even from rich libraries. These are the considerations that give me occasion to publish the present article.

Its subject is constructions on the sphere, performed only with the aid of a spherical compass.

1. Given a sphere, we can easily, by means of construction on the plane, find its radius and the spherical radius, equal to a quadrant, i.e. to a quarter of the circumference of a great circle.

To construct a given angle on the surface of the sphere at a given point \(P\) on a given arc \(PF\), we describe from the given point, as from a pole, with spherical radius equal to a quadrant, the circumference of the great circle \(FLG\); having constructed on the plane in a circle of radius equal to the radius of the sphere, with the center the given angle \(A\), we lay off the corresponding arc \(FK\) on the circumference of the great circle from the point \(F\); drawing through \(P\) and \(K\) the circumference of a great circle, we obtain the angle \(FPK\), which, as is known, is equal to the arc \(FK\) and, consequently, to the given angle \(A\).

Fig. 1.

2. The construction of triangles on a sphere is carried out, in most cases, analogously to the construction in the plane; therefore only the following constructions are of interest: a) a triangle from three angles, b) a right triangle from a leg and the opposite angle, an oblique triangle c) from two sides and the angle opposite one of them, and d) from two angles and the side opposite one of them.

In the last three cases two solutions are possible, which is why they are called cases of ambiguity.

The first problem is solved very simply: constructing—according to § 1—the three given angles \(A, B, C\) on the sphere at the point \(P\), we obtain the sides \(KG, LG, MG\) of the polar triangle \(\overline{A}\overline{B}\overline{C}\); constructing this latter by its three sides, we obtain the required triangle, as the polar triangle of \(\overline{A}\overline{B}\overline{C}\).

3. We now pass to the cases of ambiguity and first solve the problem:

To construct a right spherical triangle from a given leg and the opposite angle.

Let the leg \(c\) and the angle \(C\) be given. First of all, according to § 2, we construct the arc \(KL=\pi-C\), representing the side \(\overline{c}\) of the polar triangle.

On one side of the right angle \(A\) we lay off the arc \(AB=c\); then from \(A\), as a pole, with spherical radius \(\frac{\pi}{2}\), we draw an arc intersecting the sides of the right angle at the points \(\overline{B}\) and \(\overline{C}\); since \(\overline{B}\overline{C}=A=\frac{\pi}{2}\), the triangle \(A\overline{B}\overline{C}\) is an octant, and the points \(\overline{B}, \overline{C}\) will therefore be respectively the poles of the sides of the right angle, i.e. the vertices of the triangle polar with respect to the required triangle. Drawing from the point \(B\), as a pole, with radius \(\frac{\pi}{2}\), an arc, we obtain the side \(\overline{b}\) of the polar triangle; this arc intersects the side \(\overline{a}\) at the pole of the arc \(AB\), i.e. at the point \(\overline{C}\). If with radius \(KL\) from the point \(\overline{B}\) we cut the side \(\overline{b}\), we obtain the third vertex \(\overline{A}\) of the polar triangle \(\overline{A}\overline{B}\overline{C}\).

The point \(\overline{A}\) is the pole of the required hypotenuse \(a\); consequently, if from \(\overline{A}\), as a pole, with radius \(\frac{\pi}{2}\), an arc is drawn, then this arc will pass—

Fig. 2.

will pass through the point \(B\) and will give on \(\overline{AC}\) the point \(C\)—the vertex of the required triangle; the angle at the point \(C\), equal to \(\pi-\overline{AB}=\pi-KL\), will be the given one.

The triangles \(ABC\) and the adjacent triangle \(ABC'\) therefore satisfy the requirements of the problem.

4. We now pass to the solution and investigation of the problem:

To construct a triangle from two sides and the angle opposite one of them.

Let the sides \(a, b\) and the angle \(A\) be given.

Construct on the sphere the angle \(A\), and on one of its sides lay off the arc \(AC=b\); from the point \(C\), as pole, with spherical radius \(a\), describe the circumference of a small circle; its intersection with the second side of the angle determines the vertex \(B\) of the required triangle; we obtain its side by drawing through \(C\) and \(B\) an arc of a great circle.

Such is the general course of the construction, which, under different data, varies in details, as we shall now see. Moreover, in other cases one has to draw the arc \(CK\), perpendicular to the second side of the angle \(A\), and, as is known, this arc will be less or greater than \(\dfrac{\pi}{2}\), depending on whether the angle \(A\) is acute or obtuse.

A. Let both sides \(a\) and \(b\) be less than a quadrant of a circle.

I. Suppose first that \(a<b\), and consider the case when

1) the angle \(A\) is acute.

Depending on the magnitude of \(a\) in comparison with \(CK\), if \(a>CK\) two points of intersection \(B_1, B_2\) of the circumference of the small circle with the second side of the angle \(A\) are obtained; if \(a=CK\), both points coincide in one point \(K\), and if \(a<CK\) there will be no intersection.

In the first case, from the given elements \((a,b,A)\) we determine two triangles, one—\(AB_1C\), and the other—\(AB_2C\); moreover, the triangle \(B_1CB_2\) is isosceles, as a result of which \(B_1+B_2=\pi\).

Fig. 3.

In the second case both points \(B_1\) and \(B_2\) coincide in one point \(K\), and the required triangle \(AKC\), determined by the elements \((a=CK,\ b,\ A)\), will be right-angled, and in it

\[ c=AK,\quad B=K=\frac{\pi}{2},\quad C=ACK. \]

Finally, in the third case, the triangle cannot be constructed. It is not hard to see that the relations playing the decisive role here

\[ \tag{1} a > CK,\quad a = CK,\quad a < CK \]

coincide respectively with the usual relations

\[ \tag{2} \frac{\sin b}{\sin a}\sin A < 1,\quad \frac{\sin b}{\sin a}\sin A = 1,\quad \frac{\sin b}{\sin a}\sin A > 1; \]

indeed: here, by assumption, \(a\) and \(CK\) are less than \(\frac{\pi}{2}\), therefore with (1) the conditions

\[ \tag{3} \sin a > \sin CK,\quad \sin a = \sin CK,\quad \sin a < \sin CK; \]

are connected by mutual dependence; but, as is known,

\[ \tag{4} \sin CK = \sin A \sin b; \]

from (3) and (4) follow the corresponding conditions

\[ \tag{5} \sin a > \sin A \sin b,\quad \sin a = \sin A \sin b,\quad \sin a < \sin A \sin b, \]

coinciding with (2). Thus the geometrical meaning of these latter conditions becomes clear.

Fig. 4.

2) If the angle \(A\) is obtuse, the small-circle circumference will not intersect its sides \(ADA'\), and construction of a triangle from the elements \(\left(a < b,\ A > \frac{\pi}{2}\right)\) is impossible.

The circumference of the small circle may intersect the side of the acute angle adjacent to the given one, \(CAD'=\pi-A\); but the triangles obtained in this way, determined by the elements \(\left(a,\ b,\ \pi-A < \frac{\pi}{2}\right)\), will not answer the problem: here the sign of the inequalities (2) would indicate one or another case of solution precisely for these latter triangles, but not for the sought triangles.

II. Suppose next that \(a>b\).

The circumference of the small circle here intersects the second side of the angle \(A\) on different sides of the diameter \(AA'\); therefore,

1) If the angle \(A\) given to us is obtuse, then from the elements \(\left(a, b, A>\dfrac{\pi}{2}\right)\) we obtain a unique triangle \(AB_1C\), for which

\[ c=AB_1,\quad B_1=AB_1C,\quad C=ACB_1. \]

2) If the angle \(A\) is acute, equal to \(CAB_3\) and adjacent to the preceding angle, then the elements \(\left(a, b, A<\dfrac{\pi}{2}\right)\) also give a unique triangle \(AB_3C\), with

\[ c=AB_3,\quad B_3=AB_3C,\quad C=ACB_3. \]

Fig. 5.

The angles \(B_1\) and \(B_3\), as the angles of the isosceles triangle \(B_1CB_3\), are equal; both of them are acute, since—in view of the condition \(b<a\)—we must have

\[ B_3<B_3AC<\frac{\pi}{2}. \]

III. Finally, in the case \(a=b\), with \(A<\dfrac{\pi}{2}\), we must turn to Fig. 3: when \(a\) is increased, the point \(B_2\) will move in the direction of \(A\), and \(B_1\) in the opposite direction; therefore, when \(a=b\), the point \(B_2\) coincides with \(A\), and we obtain the unique isosceles triangle \(AB_1C\).

For an obtuse angle \(A\), as is seen from Fig. 5, it is impossible to construct an isosceles triangle for which \(B=A>\dfrac{\pi}{2}\), and consequently \(A+B>\pi\), whereas, by the condition, \(a+b<\pi\).

B. Let, next, \(a>\dfrac{\pi}{2},\ b>\dfrac{\pi}{2}\).

I. If \(a>b\), and

1) the angle \(A\) is obtuse, the perpendicular \(CK\) is greater than \(\dfrac{\pi}{2}\); therefore it will be longer than all the oblique lines. If \(a<CK\), the circumference of the small circle will intersect the side of angle \(A\) at two points \(B_1, B_2\), and we then obtain two triangles \(AB_1C\) and \(AB_2C\) satisfying the requirements of the problem; if \(a=CK\), the required triangle will be the unique right triangle \(AKC\); if \(a>CK\), the circumference of the small circle will not intersect the side \(AK\) of angle \(A\), and the

Fig. 6.

the triangle becomes impossible. In the first case the triangle \(B_1CB_2\) is isosceles, as a result of which the angles \(B_1=AB_1C\) and \(B_2=AB_2C\) supplement one another to two right angles.

The conditions for the existence of two and one solutions, or for their absence, are the inequalities

\[ \tag{6} a<CK,\quad a=CK,\quad a>CK; \]

it is not hard to see that here too these inequalities are equivalent to the inequalities

\[ \tag{7} \frac{\sin b}{\sin a}\sin A<1,\quad \frac{\sin b}{\sin a}\sin A=1,\quad \frac{\sin b}{\sin a}\sin A>1, \]

encountered above.

Indeed: for the triangle \(ABC'\), given by the elements

\[ \tag{8} a'=\pi-a<\frac{\pi}{2},\quad b'=\pi-b<\frac{\pi}{2},\quad A'=\pi-A<\frac{\pi}{2}, \]

we have two rows of respectively equivalent relations

\[ \tag{9} \frac{\sin b'}{\sin a'}\sin A'<1,\quad \frac{\sin b'}{\sin a'}\sin A'=1,\quad \frac{\sin b'}{\sin a'}\sin A'>1; \]

\[ \tag{10} a'>C'K,\quad a'=C'K,\quad a'<C'K; \]

but—in view of (8)—

\[ \tag{11} \frac{\sin b'}{\sin a'}\sin A' = \frac{\sin(\pi-b)}{\sin(\pi-a)}\sin(\pi-A) = \frac{\sin b}{\sin a}\sin A; \]

on the other hand

\[ \tag{12} C'K=\pi-CK; \]

therefore, by (11), (8), and (12), the inequalities (9) and (10) pass respectively into the inequalities (6) and (7).

2) With an acute angle \(A\), the circle of spherical radius \(a\) does not intersect the side \(ADA'\), and construction of the triangle is impossible (Fig. 7).

This circle may intersect the side \(AD'A'\) of the obtuse angle \(CAD'\), but this does not answer the problem.

II. If \(a<b\), the circle of a small circle intersects the circle of the great circle \(AA'\) at two points \(B_2\) and \(B_4\), lying on different

side of the diameter \(AA'\); in this way we shall obtain (Fig. 8)

1) for an obtuse angle \(A\), the triangle \(AB_2C\),

2) for an acute angle \(A\), the triangle \(AB_4C\),

satisfying the requirements of the problem.

Both angles \(B_2\) and \(B_4\) are equal, as the angles of the isosceles triangle \(B_2CB_4\); in the triangle \(AB_2C\) the angle \(B_2\) will be greater than the obtuse angle \(A\); consequently, both angles \(B_2\) and \(B_4\) are obtuse. The acute angles adjacent to them enter into the triangles \(A'B_2C\) and \(A'B_4C\), determined by the elements \((a,\ \pi-b,\ A'=A)\).

Fig. 7.

III. If, finally, we set \(a=b\), then for \(A>\dfrac{\pi}{2}\) the point \(B_1\) in Fig. 6 will coincide with \(A\), and we shall obtain the unique isosceles triangle \(ACB_2\); for \(A<\dfrac{\pi}{2}\) an isosceles triangle with sides \(a=b>\dfrac{\pi}{2}\) and with angles \(A=B<\dfrac{\pi}{2}\) is impossible (Fig. 7).

Fig. 8.

C. Suppose further that \(a>\dfrac{\pi}{2}\), \(b<\dfrac{\pi}{2}\).

I. Let \(a>b'=\pi-b\), and, consequently, \(a+b>\pi\).

1) If, moreover (Fig. 9), the given angle \(A\) is obtuse, the perpendicular \(CK\) will be greater than \(\dfrac{\pi}{2}\) and, consequently, longer than all the obliques. If \(a<CK\), the circle of the small circle will intersect the side of the angle \(A\) in two points \(B_1\) and \(B_2\), and then two triangles \(AB_1C\) and \(AB_2C\) will be obtained, having the prescribed elements \((a,\ b,\ A)\); for \(a=CK\) there will be one triangle, \(AKC\); if \(a>CK\), no intersection of the circles of the small circle and \(AKA'\) follows, and from the given \((a,\ b,\ A)\) no triangle can be determined. The conditions for the occurrence of one of these cases are—obviously—also here equivalent to (7).

Fig. 9.

Since in the first case, by construction, the triangle \(B_1 C B_2\) is isosceles, the angles \(B_1 = A B_1 C\) and \(B_2 = A B_2 C\) supplement one another to two right angles.

2) If the angle \(A\) is acute, the circumference of the small circle will not intersect the sides of the acute angle \(A\), and the construction of the triangle cannot be carried out (Fig. 10).

Fig. 10.

II. If \(a < b'\), and consequently \(a + b < \pi\), the circumference of the small circle will intersect the circumference of the great circle \(A A'\) at two points \(B_1\) and \(B_3\), on different sides of the diameter \(A A'\); owing to this, from the elements \((a, b, A)\) there are constructed the triangles (Fig. 11)

1) \(A B_1 C\), if the angle \(A\) is obtuse,

2) \(A B_3 C\), if it is acute.

The angles \(B_1\) and \(B_3\) are equal, as angles of the isosceles triangle \(B_1 C B_3\); since, moreover, in the triangle \(A B_1 C\), by hypothesis, \(a + b < \pi\), and the angle \(A\) is obtuse, the angle \(B_1\) must be acute; likewise \(B_3 = B_1 < \frac{\pi}{2}\).

Fig. 11.

The adjacent angles to \(B_1\) and \(B_3\) are obtuse angles and enter into the triangles \(A' B_1 C\) and \(A' B_3 C\), determined by the elements \((a, \pi - b, A' = A)\).

III. For \(a = b'\), and consequently \(a + b = \pi\), and for \(A > \frac{\pi}{2}\), the point \(B_2\) (Fig. 9) coincides with \(A'\), and the required triangle will be the unique triangle \(A B_1 C\), with acute angle \(B_1\), equal to \(\pi - A\); for \(A < \frac{\pi}{2}\) the circumference of the small circle (Fig. 10) will pass through the point \(A'\); the triangle will degenerate into the digon \(A C A'\).

C. Finally, let \(a < \frac{\pi}{2}, \quad b > \frac{\pi}{2}\).

I. Let \(a < b' = \pi - b\), and consequently \(a + b < \pi\).

1) If the angle \(A\) is acute, the perpendicular \(C K\) will be less than \(\frac{\pi}{2}\) and, consequently, shorter than all oblique lines. If \(a > C K\), the circumference of the small circle will intersect the second side of the angle at two points \(B_1\) and \(B_2\), as a result of which two triangles \(A B_1 C\) and

Constructing Triangles on a Sphere

\(AB_2C\), possessing the prescribed elements \((a,\ b,\ A)\); if \(a = KC\), one right triangle \(AKC\) will be obtained; if, finally, \(a < CK\), the circle of the small circle will not intersect the second side of angle \(A\), and there will be no triangle.

Since the triangle \(B_1CB_2\) is isosceles, the angles \(B_1 = AB_1C\) and \(B_2 = AB_2C\) supplement one another to two right angles.

2) If angle \(A\) is obtuse, the two circles will not intersect, and a triangle with elements \((a,\ b,\ A > \frac{\pi}{2})\) cannot be constructed (Fig. 13).

Fig. 12.

II. Let \(a > b'\), and, consequently,

\[ a + b > \pi. \]

The circle of the small circle intersects the circle of the great circle on both sides of the diameter \(AA'\), owing to which

1) for \(A > \frac{\pi}{2}\) one triangle \(AB_2C\) is obtained,

2) for \(A < \frac{\pi}{2}\) likewise one triangle \(AB_4C\),

which satisfy the corresponding problem; moreover the angles \(B_2 = AB_2C\) and \(B_4 = AB_4C\) are equal, and both of them, as is easy to see, are obtuse.

Fig. 13.

III. If \(a = b'\), and, consequently, \(a + b = \pi\), then for acute angle \(A\), as is seen from Fig. 12, the point \(B_1\) coincides with \(A'\), and one triangle \(AB_2C\) is obtained with obtuse angle \(B_2\), equal to \(\pi - A\); for obtuse angle \(A\) the circle of the small circle (Fig. 14) will pass through the point \(A'\), and the triangle with elements \((a = \pi - b,\ b,\ A)\) will degenerate into the two-angle figure \(ACA'\).

Fig. 14.

Thus here we obtain a complete geometrical solution of the problem under investigation.

5. Before proceeding to the geometric solution of the last problem, we must establish certain properties of the spherical biangle that we shall need.

On the side of a biangle with angle \(B\), smaller than \(\dfrac{\pi}{2}\), set off an arc

\(BC=a\), smaller than \(\dfrac{\pi}{2}\), and denote \(a'=B'C\); then from the point \(C\), as from a pole, with radius \(\dfrac{\pi}{2}\), intersect the second side of the biangle at the point

\(L\); then the arc \(CL=\dfrac{\pi}{2}\). Next construct the arc \(CK\), perpendicular to the side \(BLB'\), and, setting off \(KB''=KB\), draw the inclined arc \(CB''\), equal to the arc \(a\); then the angle \(B''=B\).

Taking on the arcs \(BK\), \(KB''\), \(B''L\), \(LB'\) the points \(W, X, Y, Z\), denote by the same letters the angles \(BWC\), \(BXC\), \(BYC\), \(BZC\), and call their adjacent angles respectively \(W', X', Y', Z'\).

Let us also recall that in a spherical triangle an exterior angle is greater or smaller than the interior angle not adjacent to it but adjoining the same side, according as the sum of the other two sides is less or greater than \(\pi\).

Since, by hypothesis, the angle \(B\) is acute, \(CK\) must be less than \(\dfrac{\pi}{2}\); hence—by the hypothesis and on the basis of known theorems—

\[ \tag{13} \frac{\pi}{2}>CB>CW>CK<CX<CB''<CY<CL=\frac{\pi}{2}<CZ<CB'; \]

keeping this in mind, let us investigate within what limits the angles \(W, X, Y, Z\) vary.

Fig. 15.

The angle \(W\) is equal to \(\dfrac{\pi}{2}\) when the point \(W\) coincides with \(K\), and is close to \(\pi-B\) when \(W\) approaches \(B\); thus

\[ \pi-B>W>\frac{\pi}{2}; \]

in exactly the same way, the angles \(X\) and \(Y\) satisfy the inequalities

\[ \frac{\pi}{2}>X>B,\qquad B>Y>L; \]

by virtue of (13), \(CL+CZ>\pi\); therefore we shall have

\[ L<Z<B. \]

If we compare all these inequalities, we obtain

\[ \pi-B>W>\frac{\pi}{2}>X>B>Y>L<Z<B; \tag{14} \]

hence it is seen that the angle \(W\), with vertex on \(BK\) and subtending the arc \(a\), will be obtuse; likewise the angles \(X,Y,Z\), arranged with vertices on \(KLB'\), will be acute and all greater than \(L\); moreover, in both directions from the point \(L\), the angles \(Y\) and \(Z\) will increase until the points \(Y\) and \(Z\) come into the position—the one \(B''\) and the other \(B'\), where \(B''=B'=B\). We thus find that

I. On the side \(BLB'\) of the lune \(BB'\), when \(B<\dfrac{\pi}{2}\) and \(BC<\dfrac{\pi}{2}\), there is no point that could be the vertex of an angle subtending the arc \(BC\) and less than \(L\).

For the adjacent angles \(W',X',Y',L',Z'\), subtending the arc \(a'=B'C\), by virtue of (14) we obtain

\[ B<W'<\frac{\pi}{2}<X'<\pi-B<Y'<L'>Z'>\pi-B, \tag{15} \]

i.e. \(L'\) proves to be the greatest angle, and \(B\) the least; hence

II. On the side \(BLB'\) of the lune \(BB'\), when \(B<\dfrac{\pi}{2}\) and \(B'C>\dfrac{\pi}{2}\), there is no point that could be the vertex of an angle subtending the arc \(B'C\) and less than \(B\).

The pole of the arc \(CK\) lies on the circle \(BLB'\); on the other side, the same pole lies on the circle described from the point \(C\), as from a pole, with radius \(\dfrac{\pi}{2}\); the point of intersection of both circles, i.e. the point \(L\), will therefore be the pole of the arc \(CK\); in this case \(KL=\dfrac{\pi}{2}\) and \(L=CK\); thus

III. In a spherical lune the least angle \(L\), subtending the arc \(BC\), is equal to the spherical distance of the point \(C\) from the other side of the lune.

Subtracting from the semicircle \(BLB'\) the arc \(KL\), in accordance with the preceding—equal to \(\dfrac{\pi}{2}\)—we obtain \(BK+LB'=\dfrac{\pi}{2}\); but since

\[ KL=KB''+B''L=\frac{\pi}{2}, \]

and, by construction, \(BK=KB''\), it follows that \(B''L=LB'\), i.e.

IV. The point \(L\) bisects the arc \(B'B''\).

We may thus say that

V. In the biangle \(BB'\), with \(B\) and \(BC\) less than \(\frac{\pi}{2}\), the smallest angle, subtended by the arc \(BC\), has its vertex at the midpoint of the arc \(B'B''\).

6. Passing to the solution of the last problem:

To construct a triangle from two angles and the side opposite one of them,

we take on the sphere a biangle with angle \(B\), and on one of its sides lay off an arc \(BC\), equal to \(a\); then draw the arc \(CK\), perpendicular to the second side of the biangle, and the arc \(CL=\frac{\pi}{2}\); finally lay off \(B''K=BK\) and join \(B''\) to \(C\) by an arc of a great circle; then it will turn out that

\[ CB''=CB=a,\quad KB''C=KBC=B. \]

For given \(a\) and \(B\), the arc \(CK\) and the angle \(L\) equal to it are completely determined.

By the known leg \(CK\) and the given angle \(A\)—according to § 3—we construct a right triangle, whose hypotenuse will be the second side of the required triangle.

Such, in general outline, is the course of the construction; this process assumes a somewhat different character under different conditions, as we shall see below.

Fig. 16.

A. We first consider the case when both given angles are acute.

If \(B\) is an acute angle, the arc \(CK\) and the angle \(L\) equal to it are less than \(\frac{\pi}{2}\); in this case, on the basis of inequalities (14), one may assert that, for \(A>B\), the required vertex of the right triangle must be situated between \(K\) and \(B'\), and for \(A<B\) this vertex may lie, first, between \(B''\) and \(L\), and secondly, between \(L\) and \(B'\).

I. Let the angle \(A<B\), and moreover

1) the side \(a\) is less than a quarter of the circumference of a great circle.

In this case, if the given angle \(A\) is greater than \(L\), one may—according to § 3—construct, on \(CK\) and \(A\), two right triangles \(A_2'KC\) and \(A_2KC\), with the vertex \(A_2\) lying between \(B'\) and \(L\); the oblique triangle \(A_2BC\) will have elements \((A_2=A, B, a)\) and will be one of those required. Laying off \(AK_1=KA_2'\) and drawing through \(A_1\) and \(C\) an arc of a great circle, we obtain in the isosceles tri-

The quadrilateral \(CA_1A'_2\) satisfies \(A_1=A'_2=A\) and \(CA_1=CA'_2\); the vertex \(A_1\), according to what was said above, will lie on the arc \(B''L\). The triangle \(A_1BC\) will be the second triangle determined by the elements \((A_1=A,\,B,\,a)\); if we denote \(A_1C=b_1,\ A_2C=b_2\), then, as is seen from the drawing, \(b_1\) and \(b_2\) supplement one another to a semicircumference. Thus, for \(A>L\), we obtain two triangles \(A_1BC\) and \(A_2BC\) satisfying the requirements of the problem; for \(A=L\) there is obtained the unique right-angled triangle \(LBC\), having the given elements \((A=L,\,B,\,a)\), and finally, for \(A<L\), by I §5, the construction of the triangle is impossible.

The occurrence of one of the three cases is determined by the existence of the relations

\[ A>L,\qquad A=L,\qquad A<L; \tag{16} \]

since here the angles \(A\) and \(L\) are acute, we have, respectively,

\[ \sin A>\sin L,\qquad \sin A=\sin L,\qquad \sin A<\sin L; \tag{17} \]

but, by III §5, \(\sin L=\sin CK=\sin B\sin a\); therefore (17) pass directly into the known inequalities

\[ \frac{\sin B}{\sin A}\sin a<1,\qquad \frac{\sin B}{\sin A}\sin a=1,\qquad \frac{\sin B}{\sin A}\sin a>1. \tag{18} \]

The construction of Fig. 16 gives us the possibility of making several important remarks.

By construction \(A_1B''=A'_2B\); as arcs of vertical angles, \(A_2B'=A'_2B\); therefore \(A_2B'=A_1B''\). If from equal arcs—according to IV §5—the arcs \(B'L\) and \(B''L\) we subtract the equal arcs \(A_2B'\) and \(A_1B''\), we obtain \(A_2L=LA_1\); hence it follows that

VI. The vertices of equal angles \(A_1\) and \(A_2\), smaller than \(B\), are equally distant from the point \(L\).

Conversely: if \(A_2L=LA_1\), then by IV §5 it will follow that \(A_1B''=A_2B'\); since \(A_2B'=A'_2B\), then \(A_1B''=A'_2B\), and consequently \(A'_2K=KA_1\); hence it follows that \(A'_2C=CA_1\), and the angle \(A'_2=A_1\); but \(A'_2=A_2\), hence \(A_2=A_1\); thus

VII. If the vertices of angles smaller than \(B\) are equally distant from the point \(L\), then these angles are equal to one another.

This theorem gives us the possibility, having found one of the vertices, say \(A_1\), of directly finding the other \(A_2\), without leaving the limits of the digon.

Further, from the construction it is clear that

VIII. The sides \(CA_1\) and \(CA_2\) of the equal angles \(A_1\) and \(A_2\) complement one another to a semicircumference.

2) If \(a\) is greater than a quadrant of the circumference, then, by II § 5, it is impossible to construct a triangle with angle \(A\) smaller than \(B\) (Fig. 17).

Fig. 17.

II. Let now \(A>B\), and (Fig. 18)

1) The side \(a\) is less than a quadrant of the circumference; in that case, in constructing a right triangle from the leg \(CK\) and the angle \(A\), the vertex \(A\) must lie on the arc \(B''K\); we then obtain the right triangle \(A_1KC\), and together with it the required triangle \(A_1BC\) with the given elements \((A_1=A,\ B,\ a)\) and with the elements to be determined

\[ b=b_1=A_1C<\frac{\pi}{2}, \quad c=A_1B, \quad C=A_1CB. \]

2) Laying off \(A_3K=A_1K\) and drawing through \(A_3\) and \(C\) an arc of a great circle, we obtain the isosceles triangle \(A_1CA_3\), for which

\[ A_3C=A_1C=b_1,\quad A_3=A_1=A; \]

at the same time we shall have, for

\[ a'>\frac{\pi}{2}, \]

the triangle \(A_3B'C\), determined by the elements \((A_3=A,\ B'=B,\ a'>\frac{\pi}{2})\) and having the remaining—sought—elements

\[ b=b_1=A_3C,\quad c=A_3B',\quad C=A_3CB'. \]

Fig. 18.

complementing to a semicircumference, are included in the triangles \(A_1'BC\) and \(A_3'B'C\), given respectively by the elements \((A,\ \pi-B,\ a<\frac{\pi}{2})\) and \((A,\ \pi-B,\ a'>\frac{\pi}{2})\).

III. Finally, when \(A=B\) we obtain (Fig. 18) the unique isosceles triangle \(BCB'\), if \(a<\frac{\pi}{2}\); but if \(a>\frac{\pi}{2}\), the construction is impossible, since we assume the sides and angles of the triangle to be less than \(\pi\).

We have just given a geometric solution of the problem under the assumption that both given angles are acute. Now we must consider the question in those cases when one or both angles are obtuse.

First of all, let us agree to call two lunes that complement one another to a hemisphere adjacent; in adjacent lunes the angles complement one another to two right angles.

B. Let both given angles \(A\) and \(B\) be obtuse.

For the lune with angle \(B\) we take the adjacent lune, whose angle \(B'=\pi-B\) will be acute.

I. Let first \(A>B\) and, consequently, \(\pi-A=A'<B'\).

1) Let then \(a>\dfrac{\pi}{2}\), and hence \(a'<\dfrac{\pi}{2}\); drawing \(CKC'\) perpendicular to the arc \(BB'''\), we construct from \(C'K\) and the known angle \(A'=\pi-A\), greater than \(L'=\pi-L\), according to § 3, a right triangle \(A_2C'K\); setting off \(A_1L=LA_2\) and drawing the arc \(C'A_1C\), we obtain a second right triangle with angle \(A_1'\), by virtue of VII § 6 equal to \(A_2'=A'\). Then, as is easy to see, the triangles \(A_1BC\) and \(A_2BC\) with elements \((A_1=A_2=A,\ B,\ a)\) will be the required ones, and here \(A<L\). If \(A=L\), the unique right-angled triangle \(LBC\) is obtained, and if \(A>L\) the construction is impossible.

Fig. 19.

For \(A<L\), as follows from VIII § 6, \(b_1+b_2=\pi\).

It is not difficult to see that here too the conditions \(A<L,\ A=L,\ A>L\) are equivalent to the conditions (18).

2) If \(a<\dfrac{\pi}{2}\) and, consequently, \(a'>\dfrac{\pi}{2}\), the construction of a triangle with obtuse angle \(A\), greater than \(B\), is impossible, as is seen from the drawing (Fig. 20) and from II § 5.

Fig. 20.

II. Let, further, \(A<B\).

Since here, in the adjacent lune, \(A'>B'\), we obtain, for given \(a'<\dfrac{\pi}{2}\), one triangle \(A_2BC'\), and for \(a=B'''C'>\dfrac{\pi}{2}\) one triangle \(A_4B'''C'\); hence we have

1) for \(a>\dfrac{\pi}{2}\), there is one required triangle \(A_2BC\), having the elements \((A_2=A,\ B,\ a)\), and

Fig. 21.

2) for \(a'<\dfrac{\pi}{2}\), likewise one triangle \(A_4B'''C\) with the given elements \((A_4=A,\ B'''=B,\ a')\).

It is clear that in these triangles the sides opposite the angles \(B\) are equal to one another and greater than \(\dfrac{\pi}{2}\).

III. When \(A=B\), we obtain (Fig. 21) one isosceles triangle \(BB''C\), if \(a>\dfrac{\pi}{2}\); if, however, \(a'\), and consequently also \(b\), is less than \(\dfrac{\pi}{2}\), the triangle is impossible, since it would violate the known theorem that the sum of two sides and the sum of the angles opposite them are uniquely related with respect to \(\pi\).

C. Let now the angle \(A\) be obtuse and \(B\) acute.

I. Let \(A'=\pi-A<B\), and consequently \(A+B>\pi\).

Fig. 22.

1) If \(a>\dfrac{\pi}{2}\), then, to construct a right triangle from the leg \(CK\) and the known angle \(A'\), greater than \(\pi-L\), we obtain the vertex \(A_1\); laying off \(A_2L=LA_1\), we obtain—according to VII § 6—the vertex of a second right triangle with the same leg and the same opposite angle; then the triangles \(A_1BC\) and \(A_2BC\), in which \(A_1=A_2<L\), will be the required ones. If \(A=L\), there results the unique right-sided triangle \(LBC\), and for \(A>L\) the construction is impossible.

In the first case—according to VIII § 6—the sides \(b_1\) and \(b_2\) complete one another to a semicircumference.

Fig. 23.

2) If \(a<\dfrac{\pi}{2}\), it is impossible to construct an angle \(A\) standing on the arc \(a\) and, by hypothesis, greater than \(\pi-B\): its vertex can lie neither on the arc \(KB'\), since it is obtuse, nor on \(BK\), since then it would be less than \(\pi-B\).

II. Let then \(A'=\pi-A>B\), and hence \(A+B<\pi\). Along the leg \(CK\) and the opposite angle \(A'\) we construct two right triangles \(A_1KC\) and \(A_3KC\), their vertices being required to lie, one on the arc \(KB''\) and the other on \(KB'\); then the required triangles will be obtained:

1) \(A_1BC\), if the side \(a>\dfrac{\pi}{2}\) is given;

2) \(A_3B'C\), if \(a'<\dfrac{\pi}{2}\) is given;

in both triangles the sides \(b_1\) are equal to one another and both are less than \(\dfrac{\pi}{2}\).

Fig. 24.

III. If \(A'=B\) and, consequently, \(A+B=\pi\), then, for \(a>\dfrac{\pi}{2}\), one triangle \(B''BC\) is obtained (Fig. 24); but if \(a'<\dfrac{\pi}{2}\), the construction cannot be carried out. Indeed: if \(a'<\dfrac{\pi}{2}\) is given, the side \(b\)—by virtue of that cited in III B of the theorem of spherical geometry—will be greater than \(\dfrac{\pi}{2}\); consequently it (Fig. 24) must be situated in the angle \(BCL\); but in that case the angle formed by it, resting on the arc \(a'\), will be acute, contrary to the assumption.

D. Finally let the angle \(A\) be acute, and the angle \(B\) obtuse.

I. Let \(A<B'=\pi-B\), and consequently \(A+B<\pi\).

1) For \(a<\dfrac{\pi}{2}\), along the leg \(CK'\) and the opposite angle \(A>L\), we construct the right triangle \(A_1CK'\); setting off \(LA_2=A_1L\) and drawing the arc \(A_2C\), we obtain a second such triangle \(A_2CK'\); then the triangles \(A_1BC\) and \(A_2BC\) with elements \((A_1=A_2=A,\ B,\ a)\) will be the required ones; moreover, according to VIII § 6, \(b_1+b_2=\pi\). For \(A=L\) one triangle \(LBC\) is obtained, and for \(A<L\) the construction cannot be made.

Fig. 25.

2) Likewise one cannot construct a triangle if in it \(a>\dfrac{\pi}{2}\) is given; indeed (Fig. 26): if the vertex of the angle \(A\), resting on the arc \(a\), lay on \(BK\), this angle would be acute, but greater than \(\pi-B\), which is contrary to the assumption; if the vertex \(A\) lay on the arc \(B'''K\), the angle would be obtuse, which is likewise not correct.

II. Let \(A > B' = \pi - B\), or, equivalently, \(A + B > \pi\). Constructing from \(CK'\) and \(A\), in the adjacent lune, the points \(A'_2\) and \(A'_4\), we obtain (Fig. 27):

Fig. 26.          Fig. 27.

1) for the given \(a > \dfrac{\pi}{2}\), the triangle \(A_2BC\), and

2) for the given \(a' < \dfrac{\pi}{2}\), the triangle \(A_4B'''C\),

having the given angles \(A_2 = A_4 = A\), \(B = B'''\), and side \(a\) or \(a'\).

III. If \(A = B' = \pi - B\) and, therefore, \(A + B = \pi\), then we obtain (Fig. 25), for \(a < \dfrac{\pi}{2}\), the triangle \(B''CB\); for \(a > \dfrac{\pi}{2}\), as we see in (Fig. 26), on the arc \(BKB''\) it is impossible to find the vertex of an angle equal to \(\pi - B\); therefore the construction of the triangle is not feasible.

Thus the last problem is solved in all possible cases.

7. An indication that the problem of § 4 admits two solutions is found in J. Müller (1436–1476), or, as he was usually called, in Regiomontanus, in his posthumous work “De triangulis omnimodis libri quinque,” published in 1533; I did not have this edition at my disposal, but I have a later edition by D. Santbech of the same book, belonging to the second half of the sixteenth century and bearing the title: “Joannis Regiomontani mathematici praestantissimi de triangulis planis et sphaericis libri quinque etc.” In this edition we find¹ a drawing explaining the origin of the double solution.

A systematic investigation of the cases of double solution, but by a trigonometric method, was published in 1756 by G. Heinsius (1709–1769) in “Acta Eruditorum”; but I have not succeeded in obtaining this edition.

The first drawing explaining the origin of the ambiguity in the problem of §6 we find in A. Cagnoli (1743–1816), in the treatise “Trigonometria piana e sferica,” published in 1786.²

A geometrical investigation of the problem of §5 is found³ in P. Lenthéric († 1849).

When the preceding had already been typeset and partly printed, I received from the library of Kazan University “Astronomie théorique et pratique” by J. B. J. Delambre (1749–1822); there⁴ I found indications of the properties of the lune that I have given above in Theorems I and III of §5 and VI, VII, and VIII of §6.

1. P. 111. ↩

2. I have the second French edition of 1808; see pp. 322–323, fig. 54. ↩

3. “Nouvelles Annales de Mathématiques,” vol. II, 1843, p. 32. ↩

4. Vol. I, p. 198. ↩