sume sin C=cos B, sin B=cos C, and likewise tang B=cot C, tang C=cot B. This being fixed, the different problems concerning right-angled triangles are all reducible to the four following cases: FIRST CASE. XXXVIII. Given the hypotenuse a, and a side b, to find the other side and the acute angles. For determining the angle B, we have (Art. 31.) the proportion ab: R : sin B. Knowing the angle, we shall also know its complement 90°-B=C; we might also find C directly by the proportion a : b:: R: cos C. Hav As to the third side c, it may be found in two ways. ing found the angle B, we can either (Art. 32.) form the proportion R cot B::b: c; or the value of c may be obtained directly from the equation c2=a2—b3, which givesc=√(a2—b2), and consequently log c=log(a+b)+¦log(a−b). SECOND CASE. XXXIX. Given the two sides b and c of the right angle, to find the hypotenuse a, and the angles. We We shall have the angle B (Art. 32.) from the proportion cb::R: tang B. Next we shall have C-90°-B. might also find C directly by the proportion b: c::R: tang C. Knowing the angle B, we shall find the hypotenuse by the proportion sin B: R::b: a; or a may be obtained directly from the equation a=√(b2+c2); but as b2+c2 cannot be decomposed into factors, this expression is incommodious in calculating with logarithms. r XL. THIRD CASE. Given the hypotenuse a, and an angle B, to find the other two sides b and c. : Make the proportions R sin B::a:b, R: cos B : : a : c ; they will give the values of b and c. As to the angle C, it is equal to the complement of B. FOURTH CASE. XLI. Given, a side b of the right angle, with one of the acute angles, to find the hypotenuse and the other side. Knowing one of the acute angles, we shall likewise know the other; hence we may look upon the side b and the opposite angle B as given. To determine a and c, we shall then have the proportions sin B Rb: a, R: cot Bb: c. SOLUTION OF RECTILINEAL TRIANGLES IN GENERAL. Let A, B, C be the three angles of a proposed rectilineal triangle; a, b, c, the sides which are respectively opposite them: the different problems which may occur in determining three of these quantities by means of other three, will all be reducible to the four following cases : FIRST CASE. XLII. Given the side a and two angles of the triangle, to find the two other sides b and c. Two of the angles being known will give us the third; then the two sides b and c will result from the proportions (Art. 33.). sin A sin B::a: b, sin A sin C::a: c. SECOND CASE. XLIII. Given the two sides a and b, with the angle A opposite to one of them, to find the third side c and the other two angles B and C. The angle B may be had by the proportion ab sin A: sin B. Let M be the acute angle whose sine is b sin A a ; from the va lue of sin B, we may either take B=M, or B=180°—M. This ambiguous solution will not occur, however, except we have at once A an acute angle and b7a. If the angle A is obtuse, B cannot be so; hence we shall have but one solution; and if, A being acute, we have ba, there will equally be only one solution, because in that case we shall have M▲ A, and by making B=180°--M, we should find A+B7180°; which it cannot be. Knowing the angles A and B, we shall also know the third angle C. Then we shall obtain the third side c by the proportion this value will not admit of being computed by logarithms, except by help of an auxiliary angle M or B, which brings it back to the foregoing solution. THIRD CASE. XLIV. Given two sides a and b, with their included angle C, to find the other two angles A and B, and the third side c. Knowing the angle C, we shall likewise know the sum of the other two angles A+B=180°--C, and their half-sum (A+B)=90°——1C. Next we shall compute the half-difference of these two angles by the proportion (Art. 36.) a+b: a—b: : tang (A+B) or cot i̟C : tang ¦(A—B), · in which we consider a 7b, and consequently A7B. Having found the half difference, by adding it to the halfsum, (A+B), we shall have the greater angle A; by subtracting it from the half-sum, we shall have the smaller angle B. For, A and B being any two quantities, we have always A=}(A+B)+}(A—B), B={(A+B)—}(A—B). Knowing the angles A and B, to find the third side c, we have the proportion sin A: sin C::a: c. XLV. In trigonometrical calculations, it often happens that two sides a and b and the included angle C, are known by their logarithms; in that case, to avoid the trouble of seeking the numbers which correspond to them, we need only seek the angle by the proportion b: a :: R: tang . The angle will be greater than 45°, since we supposed a7b; subtract from 45° therefore, and form the proportion R: : tang (4—45) :: cot 1C: tang † (A—B) ; from which, as formerly, the value of (A-B) may be de termined, and afterwards that of the two angles A and B. This solution is founded on the property, that, tang(45°-q} R❜tang -Retang 45° hence tang (4-45°)= a R and tang 45°=R; tang (p-45°) :: cot C: tang (A—B). As for the third side c, it may be found directly by means cos C a+b2-c2 of the equation R which gives But this value is inconvenient for calculating with logarithms, unless the numbers which represent a, b, and cos C are very simple. We may observe that the value of c might also be put under these two forms: c = C +(a-b)2. 3], which is easily verified by means of the formulas R2 sin2 | C = { R2— R cos C, cos2 | C={R2 + R cos C. These values will especially be useful, if it is required to compute c with great precision, the angle C and the line a- b being at the same time very small. The latter value shows that c might be the hypotenuse of a right-angled triangle, formed with the cos C R and (a--b)- ; a truth of which we may convince ourselves by this very simple construction. Let CAB be the proposed tri- E angle, in which are known the two sides CB-a, CA=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 DAF, we shall in like manner have AF: DA:: sin D: R. And substituting the values DA=DC+CA=a+b, AE=CE-CA-a-b, DC, we shall have 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 CBD=C, 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) cos C R sin C R and BF (a-b)· This construction might therefore supply the place of Art. 36. FOURTH CASE. XLVI. 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 AR2 bc ; and the other two angles may be de termined in the same way. But a different solution may be obtained by a formula more commodious for computing with logarithms. Recurring to the formula R-R cos A=2sin2 A, and substituting in it the value of cos A, we shall have |