APPENDIX I
Fig. 152.
Fig. 152.
Natural Trigonometric Functions. Consider the angleDAE,Fig. 152.From any point on the line AD drop a line perpendicular to the sideAEforming the right triangle ABC. Letarepresent the value or length of the sideBC; letbrepresent the value of the sideAC; letcrepresent the value of the sideAB. The ratio of the sideato the sidecis called the sine of the angleA. More concisely stated,a/c= sinA. The sine of an angle is the ratio of its opposite side to its hypotenuse, or opposite side over hypotenuse = sine of angleA= sinA. In a similar manner:
These ratios are known as natural functions of the angle because their values change with every change in the value of the angle.
The lengthening of the sides of the angle should not be mistaken for a change in the value of the angle. Draw to scale very carefully any angle and drop lines from any two points, as atBandB′,Fig. 152, which shall be perpendicular to the base line. Measure the sides of the triangles so formed and express their ratios as functions of the angleA. Comparing like functions of large and small triangle it will be seen that once an angle is known in degrees, its sine, cosine, etc., are determined irrespective of the length of sides. And, vice versa, if we know the functional values or ratios of certain sides of the right triangle formed about an angle, we have determined the value of the angle in degrees. The Table of Natural Trigonometric Functions, Appendix II, is nothing more than a compilation of these various ratios carefully figured out and placed in the form of a table to assist in the easy solution of problems having to do with the finding of certain parts of a triangle when other parts are given.
With a protractor, measure the angle A of the triangle whose sides were just measured, and compare the ratios of the sides or the functional values with those given in the Table, Appendix III, for the same angle. The larger the scale of the drawing, the greater the accuracy. By making use of the hundredths scale of the framing square together with a finely pointed pair of dividers, variation in values should not be great.
Solutions of Right Triangles.—By the solution of right triangles is meant the finding of unknown sides or angles when values of other sides and angles are known.
Example 1.—GivenA= 30 degrees,c= 24; FindB, a, b.Solution—B= 90-30 = 60 degrees. (The sum of the angles of a triangleequals 180 degrees.C= 90 degrees.)(1)a/c= sinA; whence,a=csinA.(a=ctimes sineA.)(2)b/c= cosA; whenceb=ccosA.
From the Tables, Appendix II, sin ofA, or 30 degrees, = .5. Substituting numerical values in (1),a= 12.
Again, from Tables, cosA, or 30 degrees, = .866. Substituting numerical values in (2),b= 20.784.
Arith. checkc²=a²+b²; 24² = 12² + 20.78²; 576 = 144 + 431.8; 576 = 576.
Graphic check.-—The graphic check which, it will be seen, might have been made use of as a graphic solution, consists in setting one square upon another with the angle of direction and the length of one side determined by the data given. That is, in this problem the protractor is set at 30 degrees and a length of 24 units is taken on the inclined square. The lengths of a and b are then carefully measured by taking a reading of the full inches and reading, the remaining fraction to hundredths by means of a sharp pair of dividers and the hundredths scale of the square.
Very many carpenters make use of graphic solutions such as this in determining rafter lengths. A little consideration, however, will show that it is a rather risky method of procedure unless the scale is large and the work scaled small. Graphs serve as easy checks against grave errors upon all kinds of work.
Example 2.—GivenAanda. To findB,c, andb.Solution—B= 90 degreesA.a/c= sinA;c=a/sinAb/c= cosA;b=ccosA.Substitute the numerical values and check as inExample 1.Example 3.—GivenAandb. To findB,a, andc.Solution—B= 90 degreesA.a/c= sinA;a=c× sinA.b/c= cosA;c=b/ cosASubstitute the numerical values and check as in Example i.Example 4.—Givenaandc. To findA,B, andb.Solution—sinA=a/c(That is, look in the tables, Appendix II, for the anglewhich has a sine equal to the result obtained by dividing the numericalvalue of the sideaby the value of the sidec.)B= 90 degreesA.b/c= cosA;b=ccosA.Substitute numerical values and check as in Example i.Example 5.—Givenaandb. To findA,B. andc.Solution—tanA=a/bB= 90 degreesA.a/c= sinA;c=a/ sinASubstitute numerical values and check as in Example 1.