time very small. The latter value shews that c might be the hypotenuse of a right-angled triangle, formed with the sides B A C Let CAB be the proposed tri- E angle, in which are known the two sides CB-a, C A=b, and the included angle C. From the point C as a centre, with the radius CB equal to the greater of the two given sides, describe a circle meeting the side CA produce in D and E; join BD, BE; and draw AF perpendicular to BD. The angle DBE inscribed in the semicircle is a right-angle; hence the lines AF, BE, are parallel, and we have the proportion BF: AE:: DF:: AD: : cos D: R. In the right-angled triangle DEF, we shall in like manner have AF : DA : : sin D: R. And substituting the values DA = DC+CA = a+b, AE CECA a-b, DC, we = = shall have AF (axb) sin } C BF= (a - b) cos & C R Hence AB the third side of the proposed triangle is actually the hypotenuse of the right-angled triangle ABF, the sides of triangle, we find the angle ABF opposite the side AF, and subtract from it the angle CBDC, we shall have the angle B of the triangle ABC. From which it appears that the solution of the triangle ABC, wherein are known the two sides a and b and the included angle C, is immediately reducible to that of the right-angled triangle ABF, wherein are known the two sides containing the right-angle, namely, AF = (a+b) sin C and BF (a-b) cos C This construction might, R Ꭱ = therefore, supply the place of Art. 47. FOURTH CASE. LVII. Given the three sides a, b, c, to find the three angles A, B, C. The angle A opposite to the side a is found by the formula b2+c2- a2 cos A=R. 2 bc ; and the other two angles may be de may termined in the same way. But a different solution be obtained by a formula more commodious for computing with logarithms. Recurring to the formula RR cos A= 2 sin2 A, § and substituting in it the value of cos A, we shall have a2 — b2 — c2 + 2 b c 2 sin2 AR2. 2bc R2. (a+b-c) (a − b + c) 2bc =R2. Hence b+c)). ((a+b-c) (a—b+e)` sin} A=R /((a+ For the sake of brevity, put 46 c a2 — (b — c) 2 _ 2 bc = } (a+b+c) = p, or a+b+c=2 p; we shall have a+b➡c=2p-2 c, a—b+c=2p-2b; hence sin ↓ A = R / . ((P—13) (p)). A formula which also gives the proportion bc: (p-b) (pc):: R2: sin2 A, and which it is easy to calculate by logarithms. Knowing the logarithms of sin A, we shall likewise know A, the double of which will be the angle sought. There are other formulas equally proper for solving the question. Thus, first, the formula R + R cos A2 cos A (b+c—a) (b+c+a). But still making a+b+c=2p, And this value being afterwards combined with sin A will give another formula; for having tang A = Examples of the Solution of Rectilineal Triangles. LVIII. Example 1. Suppose the height of a building AB were required, the foot of it being inaccessible. On the ground which we suppose to be horizontal or very nearly so, measure a base AD, neither very great nor very small in comparison with the altitude AB; then at D place the foot of the circle, or whatever be the instrument, with which we are to measure the angle BCE formed by the horizontal line CE parallel to AD, B A and by the visual ray directed to the summit of the building. Suppose we find AD or CE-67. 84 yards, and the angle BCE: = 45° 64: in order to find BE, we shall have to solve the right-angled triangle BCE, in which the angle C and the adjacent side CE are known. Here, according to Case 4., we shall form the proportion R: tang 45° 64′ : : 67 . 84 : BE. This logarithm corresponds to 59. 130; hence we have BE59.13 yards. To BE add the height of the instrument, which I suppose to be 1.12 yds., we shall have the required height AB 60.25 yds. = If, in the same triangle BCE we would know the hypotenuse, form the proportion cos 45° 64: R:: 67.84: BC. Note. If only the summit B of the building or place whose height is required were visible, we should determine the distance BC by the method shewn in the following example; this distance and the given angle BCE are sufficient for solving the right-angled triangle BCE, whose side increased by the height of the instrument will be the height required. LIX. Example 2. To find upon the ground the distance of the point A from an inaccessible object B, we must measure a base AD, and the two adjacent angles BAD, ADB. Suppose we have found AD588.45 yards, BAD=115° 48', and BDA= 40° 8'; we shall thence get the third angle ABD = 44° 44': ᏎᏎ and to obtain AB, we shall form the proportion sin ABD: sin ADB:: AD: AB. Hence the required distance AB=539.07 yards. If, for another inaccessible object C, we have found the angles CAD 39° 17′, ADC = 132° 83', we shall in like manner find the distance AC=1202.32 yards. LX. Example 3. To find the distance between two inaccessible objects B and C, (see the preceding figure,) we must determine AB and AC as in the last example; we shall, at the same time, have the included angle BAC=BAD-DAC.* Suppose AB has been found equal to 539.07 yards, AC=1202.32 yards, and the angle BAC = 76° 31'; to get BC, we must resolve the triangle BAC, in which are known two sides and the included angle. Now, by the third case, we have the propor tion AC + AB : AC — AB :: tang B+C 1749.73: 654.91 :: tang 61° 84' § : tang B-C : tang or 2 B-C 2 * It may happen that the four points A, B, C, D are not in the same plane; in which case, the angle BAC will no longer be the difference between BAD and DAC, and we shall require to have the value of that angle by a direct measurement. In other respects the operation will be exactly the same. sin B sin A:: AC: BC, or sin 93° 75' . 3 : sin 76° 31′ :: 1202 . 32 yds : BC 31° 90, 8 2 B+C 61° 84, 5 2 B 93° 75', 3 C = 29° 93', 7 Hence the required distance BC=1125 . 44 yds. LXI. Example 4. Three points A, B, C in the map of a country being given, it is required to determine the position of a fourth point M, the angles AMB, AMC being known, and the four points lying all in the same plane. On AB describe a segment AMDB capable of containing the given angle BMA; on AC describe another segment capable of containing the given angle AMC: the two arcs will cut each other in A and M; M will be the point D required. For the points of the arc AMDB are the only ones from which AB can be seen under an angle equal to |