 | John Farrar - 1822 - 244 páginas
...derive the following general rule for those cases where two sides and the included angle are known. The area of a triangle is equal to half the product of any two of its sides multiplied by the sine of the included angle, radius being unity. If the included angle... | |
 | John Farrar - 1833 - 276 páginas
...derive the following general rule for those cases where two sides and the included angle are known. The area of a triangle is equal to half the product of any two of its sides multiplied by the sine of the included angle, radius being unity. If the included angle... | |
 | John Farrar - 1833 - 274 páginas
...derive the following general rule for those cases where two sides and the included angle are known. The area of a triangle is equal to half * the product of any two of its sides multiplied by the sine of the included angle, radius being unity. » If the included angle... | |
 | Thomas Tate (mathematical master.) - 1848 - 284 páginas
...9., on sin A= — , AC .'. CD = AC x sin A, .'. area triangle ABC = I x AB x AC X sin A; that is, the area of a triangle is equal to half the product of any two sides by the sine of the included angle. Since a parallelogram is double the area of the triangle cut off by a diagonal,... | |
 | Thomas Grainger Hall - 1848 - 192 páginas
...therefore equal to half the product of the base into the altitude. Hence BG x AD acsm.B Я ~ Я ' or the area of a triangle is equal to half the product of any two sides multiplied by the sine of the angle included by them. 74. To find the area of a triangle in terms of... | |
 | Eli Todd Tappan - 1868 - 432 páginas
...without logarithms, by the formula (865), a? = V + c2 — 2bc cos. A. AREAS. 874. Theorem. — The area of a triangle is equal to half the product of any two sides multiplied by the sine of the included angle. Thus, the area of triangle ABC = \ be sin. A. For the... | |
 | Robert Hamilton Pinkerton - 1884 - 194 páginas
...to qb and also to -re. Denoting the 2 m area of the triangle ABO by A, we have, therefore, or, the area of a triangle is equal to half the product of any side and the pei-pendicular on that side from, the opposite angular point. Again, in the right-angled... | |
 | John Casey - 1888 - 300 páginas
...2i SECTION III. — AREA or TRIANGLE. 121. To find Expressions for the Area of a Triangle. 1°. The area of a triangle is equal to half the product of any two tides into the sine of their included angle. С ADB DEM. — Let АБСЪе the triangle, CD the perpendicular... | |
 | Alfred Hix Welsh - 1894 - 228 páginas
...the angle, it requires only the logarithms of £ s, \s — a, £ s — b, $ s — с. TRIGONOMETRY. THEOREM V. The area of a triangle is equal to half the product of any two sides multiplied by the sine of their included angle. в ^ Let Т denote the area of the triangle A I> С... | |
 | Harvard University - 1899 - 39 páginas
...altitude. THEOREM IV. The area of a parallelogram is equal to the product of its base and its altitude. THEOREM V. The area of a triangle is equal to half the product of its base and its altitude. Corollary. Two triangles having equal bases and equal altitudes are equivalent.... | |
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The area of a triangle is equal to half the product of any two of its sides multiplied by the sine of the included angle, radius being unity. 
