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" The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle of the base is perpendicular to the base, and bisects the vertical angle. "
Solid Geometry - Page 449
by Fletcher Durell - 1904 - 206 pages
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New Plane and Solid Geometry

Webster Wells - Geometry - 1908 - 336 pages
...spherical A is a rt. Z, and the A are equal by § 557. SOLID GEOMETRY— BOOK IX PROP. XIX. THEOREM 560. In an isosceles spherical triangle the angles opposite the equal sides are equal. Given, in spherical A ABC, AB = AC. To Prove ZB = Z C. Proof. 1. Draw AD an arc of a great O, bisecting BC...
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Plane [and Spherical] Trigonometry for Colleges and Secondary Schools

Daniel Alexander Murray - Plane trigonometry - 1908 - 358 pages
...less than the length of a great circle. This important proposition follows from Arts. 14 and 4. 6. (3) In an isosceles spherical triangle the angles opposite the equal sides are equal. (4) The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle...
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Spherical Trigonometry: For Colleges and Secondary Schools

Daniel Alexander Murray - Spherical trigonometry - 1908 - 132 pages
...less than the length of a great circle. This important proposition follows from Arts. 14 and 4. 6. (3) In an isosceles spherical triangle the angles opposite the equal sides are equal. (4) The arc of a great circle drawn from the vertex of an isosceles spherical triangle to the middle...
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Elements of Solid Geometry

William Herschel Bruce, Claude Carr Cody - Geometry, Solid - 1912 - 134 pages
...or symmetrical if their equal parts are arranged in reverse order. PROPOSITION XVIII. THEOREM. 783. In an isosceles spherical triangle the angles opposite the equal sides are equal. Given the spherical triangle ABC, in which AB = AC. To prove % B = 2£ C. Proof. Let AD, the arc of a great O, be drawn...
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Plane and Solid Geometry

George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 491 pages
...ratio as the circumferences or as the radii of the spheres (§ 382). PROPOSITION XIX. THEOREM 679. In an isosceles spherical triangle the angles opposite the equal sides are equal. Given the spherical triangle ABC, with AB equal to AC. To prove that Z B = Z C. Proof. Draw the arc AD of a great circle,...
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Plane and Solid Geometry

George Albert Wentworth, David Eugene Smith - Geometry - 1913 - 496 pages
...the circumferences or as the radii of the spheres i SPHERICAL POLYGONS PROPOSITION XIX. THEOREM 679. In an isosceles spherical triangle the angles opposite the equal sides are equal. Given the spherical triangle ABC, with AB equal to AC. To prove that ZB = ZG Proof. Draw the arc AD of a great circle,...
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Solid Geometry

Sophia Foster Richardson - Geometry, Solid - 1914 - 236 pages
...; each vertex is the pole of a great circle which passes through four other vertices. 413. THEOREM. In an isosceles spherical triangle the angles opposite the equal sides are equal. [Suggestion. Join the vertex to the midpoint of the base by the arc of a great circle.] 414. THEOREM....
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Constructive Text-book of Practical Mathematics, Volume 3

Horace Wilmer Marsh - Mathematics - 1914 - 272 pages
...one are equal respectively to a side and two adjacent angles of the other. X THEOREM 36 X THEOREM 37 In an isosceles spherical triangle, the angles opposite the equal sides are equal. X THEOREM 38 Write this theorem as the converse of theorem 37. X THEOREM 39 // two angles of a spherical...
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Solid Geometry

John H. Williams, Kenneth P. Williams - Geometry, Solid - 1916 - 184 pages
...the same order. 2. They are symmetrical if the equal parts occur in the reverse order. 795. THEOREM. In an isosceles spherical triangle the angles opposite the equal sides are equal. Let AB = AC. Draw AD an arc of a great O so as to bisect BC. Then A ABD and ACD are mutually equilateral,...
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Solid Geometry

Charles Austin Hobbs - Geometry, Solid - 1921 - 216 pages
...circle arc at its mid-point is equidistant from the extremities of the arc. * Proposition 336 Theorem In an isosceles spherical triangle, the angles opposite the equal sides are equal. HINT. Draw great O arc from the vertex of the A to the mid-point of the base, and consult Prop. 334....
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