When the result is found to the right of M or D, │P = N − 2│Q = N and in the same scale │ │ When the result is found to the right of M or D, │P = N − 1│Q = N + 1 and in the other scale │ │ When the result is found to the left of M or D, and│P = N − 1│Q = N + 1 in the other scale │ │ When the result is found to the left of M or D, and│P = N │Q = N + 2 in the same scale │ │ If the division is of the form (20° 30′)/(50), the result cannot be read off directly on the face of the rule. Thus, if in the above example 20° 30′ on S, is placed to agree with 50 on the right-hand scale of A, the result found on S under the R.H. index of A is 44° 30′. The required numerical value can then be found: (1) By placing the slide with all indices coincident when opposite 44° 30′ on S will be found 0·007 on A; or (2) In the ordinary form of rule, by reading off on the scale B opposite the index mark in the opening on the under-side of the rule. The above rules for the number of integers in the quotient do not apply in this case. If it is required to find the sine of an angle simply, this may be done with the slide in its ordinary position, with scale B under A. The given angle on scale S is then set to the index on the under-side of the rule, and the value of the sine is read off on B under the right index of A. Owing to the rapidly diminishing differences of the values of the sines as the upper end of the scale is approached, the sines of angles between 60° and 90° cannot be accurately determined in the foregoing manner. It is therefore advisable to calculate the value of the sine by means of the formula: Sine θ = 1 − 2 sin^2 (90 − θ)/(2). To determine the value of sin^2 (90 − θ)/(2). With the slide in the normal position, set the value of (90 − θ)/(2). on S to the index on the under-side of the rule, and read off the value _x_ on B under the R.H. index of A. Without moving the slide find _x_ on A, and read under it on B the value required. EX.—Find value of sine 79° 40′. Sine 79° 40′ = 1 − 2sin^2 5° 10′. But sine 5° 10′ = 0·0900, and under this value on A is 0·0081 on B. Therefore sine 79° 40′ = 1 − 0·0162 = 0·9838.