From the foregoing explanation of the manner of determining the trigonometrical functions of angles, the methods of solving right-angled triangles will be readily perceived, and only a few examples need therefore be given. Let _a_ and _b_ represent the sides and _c_ the hypothenuse of a right-angled triangle, and _a_° and _b_° the angles opposite to the sides. Then of the possible cases we will take (1.) Given _c_ and _a_°, to find _a_, _b_, and _b_°. The angle _b_° = 90 − _a_°, while _a_ = _c_ sin _a_° and _b_ = _c_ sin _b_°. To find _a_, therefore, the index of S is set to _c_ on A, and the value of _a_ read on A opposite _a_° on S. In the same manner the value of _b_ is obtained. EX.—Given in a right-angled triangle _c_ = 9 ft. and _a_° = 30°. Find _a_, _b_, and _b_°. The angle _b_° = 90 − 30 = 60°. To find _a_, set R.H. index of S to 9 on A, and over 30° on S read _a_ = 4·5 ft. on A. Also, with the slide in the same position, read _b_ = 7·8 ft. [7·794] on A over 60° on S. (2.) Given _a_ and _c_, to determine _a_°, _b_°, and _b_. In this case advantage is taken of the fact that in every triangle the sides are proportional to the sines of the opposite angles. Therefore, as in this case the hypothenuse c subtends a right angle, of which the sine = 1, the R.H. index (or 90°) on S is set to the length of _c_ on A, when under _a_ on A is found _a_° on S. Hence _b_° and _b_ may be determined. (3.) Given _a_ and _a_°, to find _b_, _c_, and _b_°. Here _b_° = (90 − _a_°), and the solution is similar to the foregoing. (4.) Given _a_ and _b_, to find _a_°, _b_°, and _c_. To find _a_°, we have tan. _a_° = _a_/_b_, which in the above example will be (4·5)/(7·8) = 0·577. Therefore, placing the slide so that the indices of T coincide with those of D, we read opposite 0·577 on D the value of _a_° = 30°. The hypothenuse _c_ is readily obtained from _c_ = _a_/(sin _a_°). THE SOLUTION OF OBLIQUE-ANGLED TRIANGLES. Using the same letters as before to designate the three sides and the subtending angles of oblique-angled triangles, we have the following cases:— (1.) Given one side and two angles, as _a_, _a_°, and _b_°, to find _b_, _c_, and _c_°. In the first place, _c_° = 180° − (_a_° + _b_°); also we note that, as the sides are proportional to the sines of the opposite angles, _b_ = (_a_ sine _b_°)/(sine _a_°) and _c_ = (_a_ sine _c_°)/(sine _a_°). Taking as an example, _a_ = 45, _a_° = 57°, and _b_° = 63°, we have _c_° = 180 − (57 + 63) = 60°. To find _b_ and _c_, set _a_° on S to _a_ on A, and read off on A above 63° and 60° the values of _b_ (= 47·8) and _c_ (= 46·4) respectively. (2.) Given _a_, _b_, and _a_°, to find _b_°, _c_°, and _c_. In this case the angle _a_° on S is placed under the length of side _a_ on A and under _b_ on A is found the angle _b_° on S. The angle _c_° = 180 − (_a_° + _b_°), whence the length _c_ can be read off on A over _c_° on S.