## First Part of an Elementary Treatise on Spherical Trigonometry |

### From inside the book

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**Solution**of Spherical Right**Triangles**. 18. To**solve**a spherical right**triangle**, two parts must be known in addition to the right angle . From the two known parts , the other three parts are to be determined , separately , by equations ... Page 17

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**solve**a spherical right**triangle**, when the hypothenuse and one of the angles are known .**Solution**. Let ABC ( fig . 2. ) be the right**triangle**, right angled at C ; and let the sides be denot- ed as in ( 427 ) . Leth and A be given , to ... Page 18

... triangles which satisfy the given values of h and A. The problem is impossible by ( 535 ) or ( 538 ) , if the ( 564 ) given value ...

... triangles which satisfy the given values of h and A. The problem is impossible by ( 535 ) or ( 538 ) , if the ( 564 ) given value ...

**solve the triangle**. Solution . by ( 561 ) , By ( 558 18 SPHERICAL TRIGONOMETRY . [ CH . II . § 11 . Page 19

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**solve the triangle**. Ans . B = 50 ° 8 ' , = 22 ° 16 ′ , b 24 ° 24 ' . a 22. Problem . To solve a spherical right triangle , when its hypothenuse and one of its legs are known . Solution . Let ABC ( fig . 2. ) be the triangle ; h the ... Page 20

... triangle ( fig . 2 ) , a = 141 ° 11 ' , and h = 127 ° 12 ' ; to

... triangle ( fig . 2 ) , a = 141 ° 11 ' , and h = 127 ° 12 ' ; to

**solve the triangle**. Ans . A = 128 ° 7 ' , B = 52 ° 22 ′ , 39 ° 6 ' . b = 25. Problem . To solve a spherical right triangle , when one of its legs and the opposite angle ...### Other editions - View all

First Part of an Elementary Treatise on Spherical Trigonometry (Classic Reprint) Benjamin Peirce No preview available - 2017 |

First Part of an Elementary Treatise on Spherical Trigonometry Benjamin Peirce No preview available - 2016 |

### Common terms and phrases

A'BC A'OC AC the perpendicular adjacent angles angle are known angles are given angles respectively equal AP and PC ar.co B'OC Cand Corollary cosec cosine of half Demonstration differs from 90 equal to 90 fall on AC Fourthly given angle given leg given sides given value greater than 90 h tang half the sum Hence hypothenuse included angle Lemma less from 90 less than 90 Let ABC fig let fall logarithm means of 496 negative obtuse opposite angle opposite side perpendicular BP perpendicular to AOC planes BOC Problem quotient right angle right tri right triangle fig right triangle PBC Scholium secant second member Secondly side BC sides and angles sides equal Solution of Spherical solve a spherical solve the triangle spherical right triangle spherical triangle ABC SPHERICAL TRIGONOMETRY substituted supplements surface ABC tang.b tangent of half Thirdly tive trian triangle ABC figs

### Popular passages

Page 69 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. D c A' D' Hyp. In triangles ABC and A'B'C', ZA = ZA'. To prove AABC = ABxAC. A A'B'C' A'B'xA'C' Proof. Draw the altitudes BD and B'D'.

Page 1 - A spherical triangle is a portion of the surface of a sphere, bounded by three arcs of great circles.

Page 69 - THEOREM. The surface of a spherical triangle is measured by the excess of the sum of its three angles above two right angles, multiplied by the tri-rectangular triangle.

Page 8 - I. The sine of the middle part is equal to the product of the tangents of the adjacent parts.

Page 8 - II. The sine of the middle part is equal to the product of the cosines of the opposite parts.

Page 30 - Any angle is greater than the difference between 180° and the sum of the other two angles.

Page 51 - The cosine of half the sum of two sides of a spherical triangle is to the cosine of half their difference as the cotangent of half the included angle is to the tangent of half the sum of the other two angles.

Page 51 - The cosine of half the sum of two angles of a spherical triangle is to the cosine of half their difference as the tangent of half the included side is to the tangent of half the sum of the other two sides.

Page 71 - ... and the sum of the angles in all the triangles is evidently the same as that of all the angles of the polygon ; hence, the surface of the polygon is equal to the sum of all its angles, diminished by twice as many right angles as it has sides less two, into the tri-rectangular triangle.