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.—Given A = 30 degrees, c = 24; Find B, a, b.
Solution—B = 90-30 = 60 degrees. (The sum of the angles of a triangle
equals 180 degrees. C = 90 degrees.)
(1) a/c = sin A; whence, a = c sin A.
(a = c times sine A.)
(2) b/c = cos A; whence b = c cos A.
From the Tables, Appendix II, sin of A, or 30 degrees, = .5. Substituting numerical values in (1), a = 12.
Again, from Tables, cos A, or 30 degrees, = .866. Substituting numerical values in (2), b = 20.784.
Arith. check c² = 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.—Given A and a. To find B, c, and b.
Solution—B = 90 degrees A.
a/c = sin A; c = a/sin A
b/c = cos A; b = c cos A.
Substitute the numerical values and check as in Example 1.
Example 3.—Given A and b. To find B, a, and c.
Solution—B = 90 degrees A.
a/c = sin A; a = c × sin A.
b/c = cos A; c = b / cos A
Substitute the numerical values and check as in Example i.
Example 4.—Given a and c. To find A, B, and b.
Solution—sin A = a / c (That is, look in the tables, Appendix II, for the angle
which has a sine equal to the result obtained by dividing the numerical
value of the side a by the value of the side c.)
B = 90 degrees A.
b/c = cos A; b = c cos A.
Substitute numerical values and check as in Example i.
Example 5.—Given a and b. To find A, B. and c.
Solution—tan A = a / b
B = 90 degrees A.
a/c = sin A; c = a / sin A
Substitute numerical values and check as in Example 1.