 | John Farrar - Logarithms - 1822 - 244 pages
...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 - Trigonometry - 1833 - 276 pages
...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 - Trigonometry - 1833 - 274 pages
...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 pages
...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... | |
 | Eli Todd Tappan - Geometry - 1868 - 432 pages
...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 altitude BD (see last figure)... | |
 | Simon Newcomb - Trigonometry - 1882 - 372 pages
...compute them for the special case d = 5, d' = 6, « = 49° 18'. Areas of Triangles. 62. THEOREM V. The area, of a triangle is equal to half the product of any two sides by the sine of their included angle. Proof. It is shown in geometry t that the area is half the base... | |
 | Robert Hamilton Pinkerton - Trigonometry - 1884 - 194 pages
...is equal 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 - Geometry - 1888 - 300 pages
...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... | |
 | Nathan Fellowes Dupuis - Geometry - 1889 - 370 pages
...one-half that of the parallelogram on the same base and altitude, .'. the area of a triangle is one-half the product of any two sides multiplied by the sine of the included angle. Or 2l6°. Theorem.— The area of any quadrangle is one-half the product of the diagonals... | |
 | Ernest William Hobson - Exponential functions - 1891 - 380 pages
...expressions \ab sin C, and żac sin B, for the area of the triangle ; the area of a triangle is therefore half the product of any two sides multiplied by the sine of the in-сluded angle. Using the expression for sin-a, found in Art. 122, gr- J(a + b + c) (6 + с - a)... | |
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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. 
