Machinery's Handbook, 31st Edition
122 COMPOUND ANGLES Example Referring to Fig. 4 : Find the compound angle C of a wedge-shaped block having known component angles A and B in sides at right angles to each other. C = compound angle in plane x - x , which is the resultant of angles A and B Solution: Let A = 47 ° 14 ′ and B = 38 ° 10 ′ ; then R tan ---------- 0.72695 = = = = R 36 ° 0.9 ′ = C 53 ° 12 ′ = In Fig. 5 is shown a four-sided block, two sides of which are at right angles to each other and to the base of the block. The other two sides are inclined at an oblique angle with the base. Angle C is a compound angle formed by the intersection of these two inclined sides and the intersection of a vertical plane passing through x - x , and the base of the block. The components of angle C are angles A and B, and angle R is the angle in the base plane of the block between the plane of angle C and the plane of angle A . Example Referring to Fig. 5 : Find the angles C and R in the block shown in Fig. 5 when angles A and B are known. Solution: Let angle A = 27 ° and B = 36 ° ; then C cot A cot 2 B cot 2 + = C 22 ° 38.6 ′ = 1.9626 2 1.3764 2 + 5.74627572 2.3971 = = = R tan B tan A tan ------ A tan R cos ------ 38 ° 10 ′ tan 47 ° 14 ′ tan ------------- 0.78598 1.0812 C tan 47 ° 14 ′ tan cos36 ° 0.9 ′ -------------- 1.0812 0.80887 ---------- 1.3367 = = = = B cot A cot ------ 36 ° cot 27 ° cot --------- 1.3764 1.9626 = R 35 ° 2.5 ′ = Example Referring to Fig. 6 : A rod or pipe is inserted into a rectangular block at an angle. Angle C 1 is the compound angle of inclination (measured from the vertical) in a plane passing through the center line of the rod or pipe and at right angles to the top sur face of the block. Angles A 1 and B 1 are the angles of inclination of the rod or pipe when viewed respectively in the front and side planes of the block. Angle R is the angle between the plane of angle C 1 and the plane of angle B 1 . Find angles C 1 and R when a rod or pipe is inclined at known angles A 1 and B 1 . Solution: Let A 1 = 39 ° and B 1 = 34 ° ; then C 1 tan A 1 tan 2 B 1 tan 2 + 0.80978 2 0.67451 2 + 1.0539 = = = C 1 46 ° 30.2 ′ = -------- 0.70131 = = = R tan 50 ° 12.4 ′ = Interpolation.— In mathematics, interpolation is the process of finding a value in a table or in a mathematical expression which falls between two given tabulated or known values. In engineering handbooks, the values of trigonometric functions are usually given only in degrees and minutes; hence, if the angle is given in degrees, minutes and seconds, the value of the function is determined from the nearest given values by interpolation. Interpolation to Find Functions of an Angle: Assume that the sine of 14 ° 22 ′ 26 ″ is to be determined. It is evident that this value lies between the sin 14 ° 22 ′ and the sin 14 ° 23 ′ . sin 14 ° 23 ′ = 0.24841 and sin 14 ° 22 ′ = 0.24813. The difference is 0.24841 - 0.24813 = 0.00028. Consider this difference as a whole number (28) and multiply it by a fraction hav ing as its numerator the number of seconds (26) in the given angle, and as its denominator 60 (number of seconds in one minute). Thus 26 ⁄ 60 × 28 = 12 (nearly); hence, by adding 0.80978 0.67451 ---------- 1.2005 = = = R 0.00012 to sin 14 ° 22 ′ we find that sin 14 ° 22 ′ 26 ″ ≈ 0.24813 + 0.00012 = 0.24825. The correction value (represented in this example by 0.00012) is added to the function of the smaller angle nearest the given angle in dealing with sines or tangents, but this correction value is subtracted in dealing with cosines or cotangents. A tan 1 B tan 1 -------
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