23-25.]

GENERAL DIRECTIONS.

41

reason is that the pupil will gain some knowledge and experience of a method that is used in other subjects, for instance, in physics and in mechanics, and that is extensively employed by engineers in solving problems in which the computations required by the other method may be overwhelmingly cumbrous.

24. General directions for solving problems. A third method of approximating to the magnitude of lines and angles may be mentioned here, for it has often to be employed in practical life. In this method the student may suppose that he possesses neither measuring instruments, drawing materials, nor mathematical tables, and thereupon he may give an off-hand estimate concerning the magnitudes required. This method also serves as a check, by showing when great arithmetical blunders are committed. The pupil is advised to use all three methods in working each practical problem in this course, and to do so in the following order:

(1) Make an off-hand estimate as to what the magnitude required may be, and write this estimate down;

(2) Solve the problem by the graphical method;

(3) Solve the problem by the slower but more accurate and reliable method of computation. There may be some interest found in comparing the results obtained by these three methods. The exercise in judging linear and angular magnitudes afforded by the first method, the practice in neat and careful drawing necessary in the second, and the training in accurate computation given by the third, will each afford some benefit to the learner.

B

25. Solution of right-angled triangles. Let ABC be a rightangled triangle, C being the right angle. In what follows, a, b, c, denote the lengths of the sides opposite to the angles A, B, C, respectively. The sides and angles of ABC are connected by the following relations:

α

These relations, however, are not needed, and are rarely used in solving triangles, for they are equivalent to (3)–(6), and few tables give secants and cosecants. See Art. 18, Note 2.

Equation (1) shows that no other element of the triangle can be derived from the two acute angles only. Each of the remaining equations, (2)–(10), involves three elements of the triangle, and at least two of these elements are sides. Hence, in order that a right-angled triangle be solvable, two elements must be known in addition to the right angle, and one of these must be a side. If any two of the elements involved in equations (2)–(10) are known, then a third element of the triangle can be found therefrom. Hence the following general rule can be used in solving right-angled triangles:

When in addition to the right angle, any two sides, or one of the acute angles and any one of the sides, of a right-angled triangle are known, and another element is required, write the equation involving the required element and two of the known elements, and solve the equation for the required element.

For example, suppose that a, c are known, and that A, B, b are required. In this case,

The quantities A, B, b can be found from these equations. If an error has been made in finding A, then B will also be wrong. Hence it is advisable to check the values found by seeing whether they satisfy relations differing from those already employed.

26, 27.]

CHECKS.

CASES.

43

For example, check formulas which may be used in this case

are:

b
α

=

tan B,

cos A.

26. Checks upon the accuracy of the computation. As already pointed out, large errors can be detected by means of the off-hand estimate and by the use of the graphical method. The calculated results in any example can be checked or tested by employing relations which have not been used in computing the results, and examining whether the newly found values satisfy these relations. An instance has been given in the preceding article. The student is advised not to look up the answers until after he has tested his results in this way. Verification by means of check formulas is necessary in cases in which the answers are not given. The testing of the results also affords practice in the use of formulas and in computation. When a check formula is satisfied it is highly probable, but not absolutely certain, that the calculated results are correct.

27. Cases in the solution of right-angled triangles. All the possible sets of two elements that can be made from the three sides and the two acute angles of a right-angled triangle are the following:

(1) The two sides about the right angle.

(2) The hypotenuse and one of the sides about the right angle. (3) The hypotenuse and an acute angle.

(4) One of the sides about the right angle, and an acute angle. (5) The two acute angles. (This case has already been referred to.)

Some examples of these cases are solved. The general method of procedure, after making an off-hand estimate and finding an approximate solution by the graphical method, is as follows:

First: Write all the relations (or formulas) which are to be used in solving the problem.

Second: Write the check formulas.

Third: In making the computations arrange the work as neatly as possible.

This last is important, because, by attention to this rule, the work is presented clearly, and mistakes are less likely to occur. The computations may be made either with or without the help of logarithms. The calculations can generally be made more easily and quickly by using logarithms.

NOTE 1. Relations (3), (4), Art. 25, may be written: a = c cos B. These relations may be thus expressed:

a = c sin A,

A side of a right-angled triangle is equal to the product of the hypotenuse and the cosine of the angle adjacent to the side.

A side of a right-angled triangle is equal to the product of the hypotenuse. and the sine of the angle opposite to the side.

NOTE 2.

a = b cot B.

Relations (7), (8), Art. 25, may be written : α= b tan 4,
These relations may be thus expressed :

A side of a right-angled triangle is equal to the product of the other side and the tangent of the angle opposite to the first side.

A side of a right-angled triangle is equal to the product of the other side and the cotangent of the angle adjacent to the first side.

EXAMPLES.

1. In the triangle ABC, right-angled at C, a = 42 ft., b 56 ft. Find B the hypotenuse and the acute angles.

Check:

b B90

42

= .7500.

56
A.

c = √a2 + b2 = √1764 + 3136.

.. A 36° 52'.2.

=

a = c cos B = 70 x cos 53° 7'.8 = 70 x .6000 42 ft.

II. Computation with logarithms.

=

*This is to be filled after the values of the unknown quantities have been found. It is advisable to indicate the given parts and the unknown parts clearly.

The work can be more compactly arranged, as follows:

45

NOTE 1. The latter form is preferable when all the parts of a triangle are required,

NOTE 2. If there is difficulty in calculating log a

log sin A in the second

form, write log sin A on the edge of a piece of paper and place it immediately beneath log a.

b

NOTE 3. The formula tan B

=

can be used instead of (2). A check

α

then is A + B = 90°. Instead of (3), one of the following formulas can be used, viz.

There is often a choice of formulas that can be used in a solution.

NOTE 4. In every example it is advisable to make a complete skeleton scheme of the solution, before using the tables and proceeding with the actual computation. In the last exercise, for instance, such a skeleton scheme can be seen on erasing all the numerical quantities in the equations that follow the logarithmic formulas.

NOTE 5. Time will be saved if all the logarithms that can be found at one place in the tables, be written at one time. Thus, for example, in the preceding exercise find log sin A immediately after A has been found.