 | Peter Nicholson - 1825 - 1046 ページ
...the sine of that angle measured in the circle ; therefore the sides of the triangle are to each other as the. sines of the opposite angles measured in the...consequently as the sines of the same angles measured in a circle whose radius is that of the tables. Hence the following proposition of such frequent use in... | |
 | William Hill (land surveyor.) - 1847 - 32 ページ
...the sine of that angle measured in the circle ; therefore the sides of the triangle are to each other as the sines of the opposite angles measured in the...Trigonometry. In any triangle, the sines of the angles measured by any one circle are proportional to the sides opposite those same angles. In our plan of... | |
 | Robert Potts - 1855 - 1050 ページ
...nearest him (from which his distance is a) in a straight line. Find the radius of the curve. 6. Shew that in any triangle the sines of the angles are proportional to the opposite sides. 8. Prove that for all values of m, cos mO + -J{- 1) sin mO is a value of {cos 0 + V(-... | |
 | Thomas Kimber - 1865 - 302 ページ
...+ g = . Account for the value of tan. A + В given by this formula, if A = В = 45°. 13. In every triangle the sines of the angles are proportional to the sides opposite to them. Find the area of the triangle whose sides are 30, 40, 50 feet. 14. Given two angles and a... | |
 | Cambridge univ, exam. papers - 1870 - 272 ページ
...particular case. /o Having given that cos 330° = ~ , find the cosine and sine of 165°. 4. Prove that, in any triangle, the sines of the angles are proportional to the sides respectively opposite to the angles, and that any side divided by the sine of the opposite angle, is... | |
 | Thomas Kimber - 1880 - 176 ページ
...В . — Account for the value of tan. A + В given by this formula, if A s= В = 45°. 18. In every triangle the sines of the angles are proportional to the sides opposite to them. Find the area of the triangle whose sides are 30, 40, 50 feet. 14. Given two angles and a... | |
 | Robert Hamilton Pinkerton - 1884 - 194 ページ
...magnitude from 0° up to 180°, but that the angle VES will always be limited in magnitude. Now, since in any triangle the sines of the angles are proportional to the opposite sides, whatever be the positions of V and E, we have the equation sinSEV_VS_5 sinSVE ~ ES... | |
 | International Correspondence Schools - 1897 - 346 ページ
...a triangle given to find the other two sides AB and C B. In Trigonometry, it is demonstrated that, in any triangle the sines of the angles are proportional to the lengths of the sides opposite to them. In other words, sin A : sin B :: BC : AC; or, sin A : sin U... | |
 | International Correspondence Schools - 1898 - 518 ページ
...method, however, is the following: In higher works on trigonometry, it has been demonstrated that, in any triangle, the sines of the angles are proportional to the lengths of the sides opposite to them. In other words, sin A : sin B::BC : AC; or, sin A : sin C:;BC... | |
 | International Correspondence Schools - 1899 - 798 ページ
...method, however, is the following: In higher works on trigonometry, it has been demonstrated that, in any triangle, the sines of the angles are proportional to the lengths of the sides opposite to them. In other words, sin A : sin B::BC:AC; or, sin A : sin C:: BC... | |
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In higher works on trigonometry, it has been demonstrated that, in any triangle, the sines of the angles are proportional to the lengths of the sides opposite to them. In other words, sin A : sin B :: BC : AC; or, sin A : sin C:: BC : AB, and sin B :... 
