In the second proportion, extend from 37° to 27 of 78° upon the sines; that extent will reach, upon the numbers, from 232 to 174 or 374, for the side AC. Given two sides and the angle included by them; to find So is the tangent of half the sum of their opposite Then the half difference, added to the half sum of the angles, gives the greater; and subtracted, leaves the less angle. ! Then, all the angles being known, find the unknown side by the first problem. NOTE the sum of the sines is to the difference of the sines of two arcs, as the tangent of half the sum of those arcs is to the tangent of half their difference in order to which, let BD, CE, be the sines of the arcs AB, AC; produce BD to the circumference at P, and produce CE till EQ be DP; to the middle point G of the are BC draw the tangent HGK, and draw CNBML parallel to it; join RH, RG, and draw ON, FB, and QL parallel to RAMK. Now it is evident, that CQ is the sum, and CF the difference, of the sines; and that GK is the tangent of half the sum AG, and GH the tangent of half the difference CG, of the two arcs AB, AC; also NM is CL, for BN NC, and BM=ML :then, in the similar triangles CQL, RMN, RGK, it will be as CQ CF :: (CQ_or) OE : (CF or) OC, And that the half sum, increased and diminished by the half difference, gives the greater and less angle respectively, is evident from the figure. And that two quantities of any kind may be found, by the same rule, from their sum and difference, may be proved thus. Let CN represent the less and NL the greater of any two quantities; and let B be the middle of the right line CL. Then it is evident, that BLBC is the half sum, and BN the half difference, as also that LB+BN-NL the greater quantity, and CB-BN-NC the less. NOTE 1. When, in this case, the triangle is right-angled, the longest side will be found by extracting the square root of the sum of the squares of the other two sides; and then the angles will be found by the first problem. NOTE 2. That instead of the tangent of the half sum, we may use the cotangent of half the given angle, which is the same thing, 1. Draw AB equal to 345, from a scale of equal parts. 2. Make the angle A equal to 37°. 20. 3. Make AC equal to 174'07, by the scale of equal parts. 4. Join B, C, and it is done. Then, the parts being measured, we have the C= 115°, the B27°, and BC232 yards. Arithmetically. Arithmetically. As sum of sides AB+AC 51907 27152259 AC 170'93 22328183 To diff. of sides AB So tang. To tang. 2 In the first proportion, extend from 519 to 171 on the line of numbers ; that extent will reach, upon the tangents, from 71°, (the contrary way, because the tangents are set back from 45°) a little beyond 45, which, being set so far back from 45, falls upon 44°4, the fourth term. In the second proportion, extend from 64° to 37 on the sines; that extent will reach, on the numbers, from 345 to 232, the fourth term required. |