One of the angles of that triangle being given equal to 47°, and the other equal to 105° 30', it will be easy to find the third angle, by recollecting that the sum of the three angles of a triangle is equal to two right angles (Geom. Theorem 15), or 180°; therefore subtracting the sum of the two given, 47° 105° 30' 152° 30' from 180°, the remainder 27° 30' is the third angle of the triangle, and is opposite the given side 500 yards. Thus by the proportion 791 yds 105 500 yds sin 27° 30': sin 470 :: 500: dist. from It. house to fort. 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.. EXERCISE. Given in an oblique angled triangle, A 75° 30' 18.5, B = 45° 16', c = 1145.3, to find the other parts of the triangle. Ans. c 59° 13′ 41′′•5, a = 1290.55, b = 946.949. We add another practical EXAMPLE 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. Theorem 13), 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 330 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 101o: sin 55° 54' :: hypoth. : height required. from which, multiplying means and dividing by the first term, substituting this value in the last equation, we have 118.6 yards, or 355.8 feet is the height of the hill. log. of 118 6. To this result the height of the eye or of the instrument should be added. 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? It must be observed that 9.85123 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 8, Geom. 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 be greater than the other given side, because in every plane triangle the greater angle is opposite the greater side. 69. 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 ABC 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. Th. 29.) A B D whence multiplying the extremes and dividing by the third term substituting in this expression for B D its value obtained above, we have 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. But A B D is the supplement of the angle B of the triangle A B C, hence substituting-cos в for cos A B D above, and changing the signs we have 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 same name 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 EXAMPLE. Suppose the three sides of a triangular plat of ground are to be 50, 60, and 70 yards, under what angle must the first two be laid out? 25003600 - 4900 cos required angle = R X ᄒ 2 X 50 X 60 = if we make R 1; hence or 20000 is the nat. cosine of the angle required. This angle will be found from the table of sines and cosines, to be 78° 27' 47".* This case might also be solved with the table of log. sines, &c., by subtracting the log. of the whole denominator from that of the whole numerator, and adding 10 the log. of R, or adding at once the arith. comp. of the denominator, the logs. of the numerator and R, rejecting 10 from the sum; in either case the result would be log. cosine of the angle required. The solution is left as an exercise for the student. 70. We now proceed to demonstrate some formulas which express relations between the different trigonometrical lines of the same arc, and The seconds are found as follows: Take the difference between the two cosines next greater and next less than yours, and also the difference between yours and the next greater, multiply the latter difference by 60 and divide the product by the former difference; the quotient will be the seconds sought. The reason appears from the following proportion. |