Arith. comp. log. sin 30o = 0.301030 (Art. 15) log. sin 102° =log.sin 78° = 9.990404 log. 1000 = 3.000000 791yds 3.291434=log.of1956 The length of the causeway must be 1956 yards. 65. This proportion is also applicable to the case where two angles and the interjacent side of a triangle are given. Such is the case in the problem at Art. 10. One of the angles of that triangle being given equal to 47°, and the other equal to 105° 30', i it will be easy to find the third angle, by recollecting that the sum of the three angles of a 1054 triangle is equal to two right angles (Geom. B. 1, Prop. 25,) or 180°; therefore subtracting the sum of the two given, 47° + 1050 30' 152° 30' from 180°, the remainder 270 30', is the third angle of the triangle, and is opposite the given side 500 yards ; hence sin 27° 30': sin 470 :: 500 : dist. from It, ho. to ft. arith. comp. log. sin 27° 30'= 0.335594 log.sin 470 - 9.864127 14 500 yds 2.898691 = log. 791.9 The distance from the light house to the fort is 791.9 yards. The importance, where any considerable degree of accuracy is required, of the method of solution by calculation, instead of that by construction, will appear from the fact, that with tolerably accurate instruments, and some care in the construction, we made the required side, which we here find to be 791.9 yards, to be 800 yards upon the scale ; thus committing an error in the construction of about 8 yards. 66. We add another practical EXAMPLE. A 054 33°20' 100 yds Involving the same case of solution combined with the solution of a right angled triangle. An observer upon a plain, desires to find the height of a neighboring hill above the level of the plain. Near the foot of the hill, let him take the angle of elevation to the top, and suppose it to be 55° 54'; then let him measure back a distance, say 100 yards, and again take the angle of elevation, which let be 33° 20'. Then in the triangle, of which 100 yards is the base, and 33° 20' is one of the angles at the base, we may have the angle at the vertex, and opposite to the given base, by observing that the exterior angle of a triangle being equal to the two interior and opposite (Geom. B. 1, Prop. 25, Cor. 6) one of the interior is equal to the exterior minus, the other interior, and therefore the angle at the vertex here is equal to 55° 54' 33° 20' = 22° 34'; then say as in the last example sin 22° 34' : sin 33° 20' : : 100 : side opp. to 33° 20'; but in the right angled triangle of which 55° 54' is the angle at the base, and the height of the hill one of the perpendicular sides, we have the proportion 1010 : sin 55° 54' : : hypoth. : height required. from which, multiplying means and dividing by the first term, sin 55° 54' x hypoth. height req. = 1010 but from the preceding proportion sin 33° 20' x 100 hypoth. or side op. 33° 20' = sin 22° 34' substituting this value in the last equation, we have sin 55° 54' x sin 33° 20' x 100 height required 1010 x sin 22° 34' : sum rejecting twice 10 = 2.073979=log. 118.6* 118.6 yards, or 355.8 feet, is the height of the hill. EXAMPLE III, 22'37 117 A 117ft 67. Let there be a A 216 feet street in which the front of a triangular block is 216 feet, and another street making an angle of 22° 37' with the first; B under what angle must a third street be laid out from the extremity A of the first, so that the front of a complete row of buildings upon it shall be 117 feet in length ? SOLUTION. 117: sin 220 37' : :216: the angle B arith. comp. of log. 117=7.931814 log. sin 22° 37' =9.584968 : log. 216 = 2.334454 sum rejecting 10=9.851236=log. sin 45° 13' 55" = =B and 180° — B-c=112° 9'5" = angle a required. It must be observed that 9.851236 is also the log. sine of the supplement of 45° 13' 55", because the sine of an arc is equal to the sine of its supplement. (Art. 15.) Hence 134° 46' 5'' is also a value of the angle B, and there are two solutions to the problem, which are both exhibited in the diagram. This case corresponds to Problem 11,B. 3, Geom. To this result the height of the eye or of the instrument should be added. If the given angle were right or obtuse, there could be but one solution, and the required angle must be acute. The same is the case if the given side opposite the given angle is greater than the other given side, because in every plain triangle the greater angle is opposite the greater side. 68. We shall next derive a formula for the solution of a triangle, when the three sides are given, and one or all of the angles required. Let A B C be any triangle; from the vertex of one of the angles a, let fall a perpendicular A D upon the side opposite. This perpendicular may fall either within or without the triangle. First, suppose that it falls within; then (Geom. B. 4, Prop. 12.) A C2 = A B2 + B C2. 2 BCX BD Also in the right angled triangle A B D, we have R: COS B: A B B D (Art. 38) whence multiplying the extremes and dividing by the third term COS B RX BD substituting in this last expression for B D its value obtained above, we have A B2+B C2 A c2 C2 COS BRX 2ABXBC an expression for the cosine of an angle in terms of the three sides of a triangle. Suppose now that the perpendicular falls without the triangle. Again, in the right angled triangle A B D R COS ABD: A B B D But A B D is the supplement of the angle B of the triangle A B C, hence COS B for cos A B D above and changing the A B substituting for B D in the second member of this equation its value found above, we have as before or employing the small letters to represent the sides opposite the angles which are expressed by the large letters of the That is to say the cosine of either angle of a triangle is equal to the sum of the squares of the two sides which contain it, minus the square of the side opposite, divided by twice the rectangle of the containing sides.. Let us apply this formula to an |