places on the same horizontal straight line, together with the distances between the stations; to find the height of the object and its distance from either station, Let AB be the object, and c, c', c", the three stations, then the triangles BCA, BC'A, BC"A, will all be right angled at A; and, therefore, to radius BA, Ac, ac', ac", will be the tangents of the angles at B, or the cotangents of the angles of elevation; hence, putting a, a, a", for the angles of elevation, x for the height of the object, and a, b, for the distances c c', c' c', we shall have B A P AC=x cot α, AC'x cot a', AC"=x cot a" Now if a perpendicular AP be drawn from A to cc", we shall have (Geom., p. 80,) from the triangle ACC' Ac2=ac22+c'c2—2 c'c, c'p; and from the triangle Ac'c" AC"ACC"c2+2 c"c'. c'P; that is, we shall have the two equations x2 cot2 ax2 cot2 a'+a2-2a. c'p x2 cot2 a"=x2 cot2 a'+b2+2b, c'p, in order to eliminate c'r, multiply the first by b, the second by a, and add, and we shall have x2(b cot2 a+a cot2 a")=(a+b) x2 cot2 a'+ab (a+b). ab (a+b) b cot2a+a cot2 a"(a+b)cot2 a'' If the three stations are equidistant, then a=b, and the expression becomes a cots a+cot a"—cot2 a The height AB being thus determined, the distances of the stations from the object are found by multiplying this height by the cotangents of the angles of elevation. Solution of Quadrantal Triangles. (124.) The rules for right-angled triangles will serve also for the solution of quadrantal triangles, or those in which one side is a quadrant. For by changing such a triangle for its supplemental or polar triangle, we shall then have to consider a right-angled triangle, of which the hypothenuse will be the supplement of the angle opposite the quadrantal side, the two perpendicular sides supplements of the other two angles of the proposed triangle, and the two oblique angles of the new triangle supplements of the oblique sides of the primitive triangle. That is, the sides of the primitive or quadrantal triangle being a, b, and c = 90°, and its angles A, B, C, the sides of the supplemental triangle will be 180°-A, 180°—B, and 180°-c, this latter being the hypothenuse; and the opposite angles will be 180°-a, 180o-b, and 90°. But the parts of a quadrantal triangle may be determined with. out the aid of the supplemental triangle. Thus let AD be the quadrantal side in the triangle ABD. Produce DB, if necessary, till DC becomes a quadrant, and draw the the arc ac, which will, obviously, measure the angle D, since D will be the pole of the arc AC, and c will be a right-angle: also the angle CAB will be the complement of the angle BAD in the proposed triangle, and the angle ABC will either be identical with ABD in the proposed, or supplemental to it, accordingly as DC exceeds, or falls short of, a quadrant; hence all the parts of the proposed triangle are easily determined from those of the right-angled triangle ABC. A B If the angle DAB is less than 90°, or than the angle dac, the side DB must, obviously, be acute; but if DAB is greater than 90°, DB will be obtuse, and conversely. Hence the angles adjacent to the quadrantal side are of the same species as the sides opposite to them. The same may be infer red from the polar triangle. It must be remarked, that the solution will be ambiguous whenever the determination of the right-angled triangle becomes ambiguous, whether we employ the polar triangle, or the triangle ABC in the above diagram. This ambiguity occurs only when the given parts in the right-angled triangle are one of the perpendicular sides, and the opposite angle to it. Let A be the middle part, then b and c will be adjacent parts, therefore, that is, rad. X sin. comp. a = tan. b × tan. comp. c, cot.c 48° 0' 9" 9.954368 Let B be the middle part, then a, b, will be opposite parts, and, consequently, that is, rad. X sin comp. в = cos b × cos comp. a; hence the angle ABD is 115° 20′ 5′′. It remains now to find a; let, therefore, в be the middle part, then a and c will be the adjacent parts; hence that is, rad X sin comp. в tan a × tan comp. c; therefore, the side DB, which is the complement of this, is 64° 34' 40." 29 2. In the triangle DAB, DA= 90°, =112° 2' 9", and AB = 67° 3' 14", to find the other parts. D Since in this example A is obtuse, DB is obtuse. In the right-angled triangle ABC we have A = 22° 29' and ab= 67° 3' 14"; let A be the middle part, then AB, AC, will be adjacent parts, and we shall have rad x sin comp. A=tan b x tan comp. €; that is, B rad cos A rad COS A = tan b cot c .. tan b = cotc therefore, the angle = 65° 27' 9", Take now a for the middle part, then A and c will be opposite parts ; hence rad x sin a= cos comp. A X cos comp. C, that is, rad sin a=sin A sin c.. sin a= sin A sin c rad and a will be acute, because the opposite angle is acute rad 10.000000 220 2 9" 9.574247 67 3 14 9.964199 sin A sin c sin a 20 12 44 9.538446; therefore BD = 1100 12' 44". As we have now to find B, take a for the middle part, then b and B will be adjacent parts, therefore |