(3.) Given the sides and the included angle, to find the other side and the remaining angles. If, for example, there are given _a_ = 65, _b_ = 42, and the included angle _c_° = 55°, we have (_a_ + _b_) ∶ (_a_ − _b_) = tan. (_a_° + _b_°)/(2) ∶ tan. (_a_° − _b_°)/(2). Then, since _a_° + _b_° = 180° − 55° = 125°, it follows that (_a_° + _b_°)/(2) = (125°)/(2) = 62° 30′. By the rule for tangents of angles greater than 45°, we find tan. 62° 30′ = 1·92. Inserting in the above proportion the values thus found, we have 107 ∶ 23 = 1·92 ∶ tan. (_a_° − _b_°)/(2). From this it is found that the value of the tangent is 0·412, and placing the slide with all indices coinciding, it is seen that this value on D corresponds to an angle of 22° 25′. Therefore, since (_a_° + _b_°)/(2) = 62° 30′, and (_a_° − _b_°)/(2) = 22° 25′, it follows that _a_° = 84° 55′, and _b_° = 40° 5′. Finally, to determine the side _c_, we have _c_ = (_a_ sin _c_°)/(sin _a_°) as before. PRACTICAL TRIGONOMETRICAL APPLICATIONS. A few examples illustrative of the application of the methods of determining the functions of angles, etc., described in the preceding section, will now be given. To find the chord of an arc, having given the included angle and the radius. With the slide placed in the rule with the C and D scales outward, bring one-half of the given angle on S to the index mark in the back of the rule, and read the chord on B under twice the radius on A. EX.—Required the chord of an arc of 15°, the radius being 23 in. Set 7° 30′ on S to the index mark in the back of the rule, and under 46 on A read 6 in., the required length of chord on B. To find the area of a triangle, given two sides and the included angle. Set the angle on S to the index mark on the back of the rule, and bring cursor to 2 on B. Then bring the length of one side on B to cursor, cursor to 1 on B, the length of the other side on B to cursor, and read area on B under index of A. EX.—The sides of a triangle are 5 and 6 ft. in length respectively, and they include an angle of 20°. Find the area. Set 20 on S to index mark, bring cursor to 2 on B, 5 on B to cursor, cursor to 1 on B, 6 on B to cursor, and under 1 on A read the area = 5·13 sq. ft. on B. To find the number of degrees in a gradient, given the rise per cent. Place the slide with the indices of T coincident with those of D, and over the rate per cent. on D read number of degrees in the slope on T. As the arrangement of rule we have chiefly considered has only a single T scale, it will be seen that only solutions of the above problem involving slopes between 10 and 100 per cent. can be directly read off. For smaller angles, one of the formulæ for the determination of the tangents of submultiple angles must be used. In rules having a double T scale (which is used with the A scale) the value in degrees of any slope from 1 to 100 per cent. can be directly read off on A.