Spherical trigonometryJ. Noon, 1736 |
Dentro del libro
Resultados 1-5 de 28
Página 7
... Interfection of the Meridian with the Hori- zon ; fuch is HBO . XXXVII . The Tropics are two fmall Circles paral- lel to the Equinoctial , and diftant ( on each Side ) there- from 23 ° 29 ′ ; they are the Limits of the Sun's great- eft ...
... Interfection of the Meridian with the Hori- zon ; fuch is HBO . XXXVII . The Tropics are two fmall Circles paral- lel to the Equinoctial , and diftant ( on each Side ) there- from 23 ° 29 ′ ; they are the Limits of the Sun's great- eft ...
Página 20
... Interfection a , a , a , a , & c . Draw the Curvelineal Arch Aa , a , a , & c . C ... After the fame manner draw the other three Arches CB , BD , DA ; fo fhall the Whole Ellipfis ACBD be compleat as required . 2 E. F. PROBLEM II . To ...
... Interfection a , a , a , a , & c . Draw the Curvelineal Arch Aa , a , a , & c . C ... After the fame manner draw the other three Arches CB , BD , DA ; fo fhall the Whole Ellipfis ACBD be compleat as required . 2 E. F. PROBLEM II . To ...
Página 34
... FM , and EHG , those Triangles are fimiliar , and fubcontrarily pofited , Now because both the Lines FM , G H , are in the famc Plane DEC ; There Therefore the common Interfection of the Plane IL , is 34 Theorems ferving to.
... FM , and EHG , those Triangles are fimiliar , and fubcontrarily pofited , Now because both the Lines FM , G H , are in the famc Plane DEC ; There Therefore the common Interfection of the Plane IL , is 34 Theorems ferving to.
Página 35
Benjamin Martin. Therefore the common Interfection of the Plane IL , is at Right Angles to both FM and GH . Hence it will be FKX KM KLq , because of the Angles EFM EHG , GKF = MKH ; therefore are the Triangles fimiliar GKF and HKM ...
Benjamin Martin. Therefore the common Interfection of the Plane IL , is at Right Angles to both FM and GH . Hence it will be FKX KM KLq , because of the Angles EFM EHG , GKF = MKH ; therefore are the Triangles fimiliar GKF and HKM ...
Página 51
... Interfection , e , be limited , then draw a Circle through e and a , by Prob . 4 . VOL . II . H 2 PRO PROBLEM IX . To graduate any leffer Oblique Circle . the Stereographical Projection . 51 . To draw a Great Circle perpendicular to other.
... Interfection , e , be limited , then draw a Circle through e and a , by Prob . 4 . VOL . II . H 2 PRO PROBLEM IX . To graduate any leffer Oblique Circle . the Stereographical Projection . 51 . To draw a Great Circle perpendicular to other.
Contenido
1 | |
2 | |
19 | |
25 | |
43 | |
49 | |
55 | |
103 | |
240 | |
246 | |
252 | |
258 | |
264 | |
275 | |
281 | |
294 | |
112 | |
124 | |
131 | |
138 | |
161 | |
166 | |
200 | |
208 | |
213 | |
221 | |
234 | |
297 | |
305 | |
316 | |
322 | |
325 | |
334 | |
341 | |
347 | |
353 | |
359 | |
366 | |
Términos y frases comunes
adjacent Angle alfo Altitude Analemma Analogy Arch Azimuth Bafe Baſe becauſe BIFH Cafe Center Chords Circles of Latitude Co-fine Co-tangent Colure Complement confequently Cufp Declination defcribe Degrees Demonftration Dial Diameter Diſtance draw E. D. THEOREM Eaft Ecliptic equal Equinoctial faid fame fhall fhew find the Angle find the Hypothenufe find the Leg find the Side firft firſt folving fuch fuppofe given the Side Globe greateſt half Sum Half-Tangent Horizon Hour-Lines Houſes Interfection laft Latitude leffer lefs Longitude meaſured Meridian muſt North Numbers Oblique Circle oppofite paffeth Parallel Perpendicular Plane recline Pofition Point Pole Prime Vertical Prob PROBLEM Projection Quadrant Radius Reclining Plane Right Afcenfion Right Line Right-angled Spherical Triangles Scheme Semicircle Sine Sine of half Sphere Spherical Angle Spherical Trigonometry Sun's Tangent of half thefe theſe thofe thoſe Tropic of Capricorn Weft Wherefore whofe
Pasajes populares
Página 75 - The three angles of a spherical triangle are together greater than two right angles and less than six right angles. Let A, B, C be the angles of a spherical triangle ; let a', b', o' be the sides of the polar triangle. Then by Art. 30, a...
Página 185 - TO THEIR DIFFERENCE ; So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES', To THE TANGENT OF HALF THEIR DIFFERENCE.
Página 186 - 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.
Página 186 - 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.
Página 186 - The sine of half the sum of two angles of a spherical triangle is 'to the sine of half their difference as the tangent of half the included side is to the tangent of half the difference of the other two sides.
Página 241 - It commences in the morning and ends in the evening, when the sun is 18° below the horizon.
Página 5 - Equinoctial (counted from the beginning of Aries) which cometh to the Meridian with the Sun or Stars, or with any portion of the Eclyptick.
Página 365 - Dialogue, adapted purpofely to the Capacities of the Youth of both Sexes ; and adorned and illuftrated with variety of Copper- Plates.
Página 75 - Side *»» is the Supplement of the Angle H, and the Angle E of the Side G D.
Página 205 - Superficies to M ; And, as two right Angles are to F, So is half the fpherical Superficies to K.