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PROBLEM IV.

721. To pass a circle through auy three points, or to find the circumcenter of any spherical triangle.

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Find the intersection point of perpendiculars erected at the midpoints of two sides.

THEOREM XX.

722. Any angle made with a side of a spherical triangle by joining its extremity to the circumcenter, equals half the anglesum less the opposite angle of the triangle.

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For X A + B + C = 24 OCA + 2% OCB + 2 * OAB,

* A + B + 4C i. * OCA =

(% OCB I X OAB)

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723. COROLLARY. Symmetrical spherical triangles are equivalent.

For the three pairs of isosceles triangles formed by joining the vertices to the circumcenters having respectively a side and two adjacent angles equal, are congruent.

THEOREM XXI.

724. When three lines mutually intersect, the two triangles on opposite sides of any vertex are together equivalent to the lune with that vertical angle.

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For DF BC, having the common supplement CD; and FA CH, having the common supplement HF; and AD = BH, having the common supplement HD;

2 ADF = ABCH, . 2 ABC + Î ADF = À ABC + 2 BCH = lune ABHCA.

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725. The Spherical Excess of a spherical triangle is the excess of the sum of its angles over a straight angle. In general, the spherical excess of a spherical polygon is the excess of the sum of its angles over as many straight angles as it has sides, less two.

THEOREM XXII.

726. A spherical triangle is equivalent to a lune whose angle is half the triangle's spherical excess.

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PROOF. Produce the sides of the Ĉ ABC until they meet again, two and two, at D, F, and H. The ĉ ABC now forms part of three lunes, whose angles are A, B, and C respectively.

But, by 724, lune with X A = Î ABC + Î ADF.

Therefore the lunes whose angles are A, B, and C, are together equal to a hemisphere plus twice Î ABC.

But a hemisphere is a lune whose angle is a straight angle,

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2 À ABC = lune whose X is (A + B + C - st.)

lune whose X is e.

727. COROLLARY I. The sum of the angles of a spherical triangle is greater than a straight angle and less than 3 straight angles.

728. COROLLARY II. Every angle of a spherical triangle is greater than je.

729. COROLLARY III. A spherical polygon is equivalent to a lune whose angle is half the polygon's spherical excess.

730. COROLLARY IV. Spherical polygons are to each other as their spherical excesses, since, by 703, lunes are as their angles.

731. COROLLARY V. To construct a lune equivalent to any spherical polygon, add its angles, subtract (n − 2) straight angles, halve the remainder, and produce the arms of a half until they meet again.

THEOREM XXIII.

732. If the line be completed of which the base of a given spi al triangle is a sect, and the other two sides of the triangle be produced to meet this line, and a circle be passed through these two points of intersection and the vertex of the triangle, the locus of the vertices of all triangles equivalent to the given triangle, and on the same base with it, and on the same side of that base, is the arc of this circle terminated by the intersection points and containing the vertex.

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Let ABC be the given spherical triangle, AC its base. Produce AB and CB to meet AC produced in D and E respectively. By 719, pass a circle through B,D,E. Let P be any point in the arc EBD. Join AP and CP. AP produced passes through the opposite point D, and CP through E, forming with DE the Ĉ PDE. Join F, its circumcenter, with P, D, and E. Since, by 678, the two angles of a lune are equal,

X PAC = st. - * PDE,
* PCA = st. - * PED,
* APC = X DPE;

.. 4 (PAC + PCA + APC) = 2 st. XS-X(PDE + PED DPE)

= 2 st. Xs – 24 FDE = a constant. (722. Any angle made with a side of a spherical triangle by joining its extremity

to the circumcenter, equals half the angle-sum less the opposite angle of the spherical triangle.)

733. In a sphere a line is said to touch a circle when it meets the circle, but will not cut it.

THEOREM XXIV.

734. The line drawn at right angles to the spherical radius of a circle at its extremity touches the circle.

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Let BD be perpendicular to the spherical radius AB. Join A with any point C in BD.

By 716, AC > AB, therefore C is without the circle.
And no other line through B, as BF, can be tangent.
For draw AEI BF. By 716, AB > AE,

E is within the circle.

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