13. Use of the Large Tables. Five-place tables are used in precisely the same manner as the small table of p. 15. Example 1. One angle of a right triangle is 42° 20' and the hypotenuse is 28 ft. 6 in. long. Find the remaining sides and the other angle. Draw a diagram to illustrate the problem, indicating the given parts. Denote the unknown parts by the letters a and b, as in Fig. 13. To find b, note that it is the side adjacent to the given angle, and that the hypotenuse is given. Hence, by (14), § 6, b = 28.5 cos 42° 20′ 28.5 × .73924 = 21.07. = H b h 42 FIG. 13 Note that a is opposite the given angle; hence by (13), § 6, a = 28.35 sin 42° 20′ 28,5 x .67344 = 19.09; the sine and the cosine of 42° 20′ being found in a table. To find ẞ, note that it is the complement of 42° 20′; hence ẞ = 77° 40'. Example 2. The perpendicular sides of a right triangle are 22 ft. 6 in. and 54 ft., respectively. Find the hypotenuse and the angles. Draw a diagram, indicating the given parts and lettering the parts to be found, as in Fig. 14. To find α, note that the given parts are the sides opposite and adjacent to it; hence by the definition of tangent, we may write tan a 22.5 54.41660. A FIG. 14 Another method is the following: By the Pythagorean theorem of plane geometry, using a table of squares and square roots or by direct calculation, Hence h = 58.5. But this method is arduous without a See Table VI. Having found α = 22° 37′ we might find h as follows. By (15), § 6, h = 54/(cos 22° 37') = 54/.92310 = 58.498. This method is no shorter than the one used above, and is open to the objection that any error made in computing a vitiates the resulting value found for h. In general, compute each unknown part from the given parts; i.e. do not use computed parts as data if it can be avoided. As in these Examples, observe the procedure suggested on p. 14, Exercises IV, in solving any triangle. EXERCISES VI.-RIGHT TRIANGLES LARGE TABLES 1. Solve the following right triangles. The hypotenuse is denoted by h, other sides by other small letters, and any angle by the capital letter corresponding to the small letter that denotes the side opposite it. 2. The base of an isosceles triangle is 324 ft., the angle at the vertex is 64° 40'. Find the equal sides and the altitude. 3. A chord of a circle is 21.5 ft., the angle which it subtends at the center is 41°. Find the radius of the circle. 4. To determine the width BA of a river, a line BC 100 rods long is laid off at right angles to a line from B to some object A on the opposite bank visible from B. The angle BCA is found to be 43° 35'. Find AB. 5. The shadow of a tower 200 ft. high is 252.5 ft. long. angle of elevation of the sun? What is the 6. Two ships in a vertical plane with a lighthouse are observed from its top, which is 200 ft. above sea level. The angles of depression of the two ships are 15° 17' and 11°22'. Find the distance between the ships. 14. Projections. The projection of a line segment AB upon a line is defined to be the portion MN of the line 7 between perpendiculars drawn to it from A and B, respectively. The or, the projection of a segment upon a given line is equal to the product of the length of the segment and the cosine of the angle the segment makes with the given line. The projections of a segment upon the coördinate axes are frequently used. If the segment makes an angle a with the horizontal, the projections on the x and y axes are, respectively, (1) Proj, AB=AB cos a, Proj, AB=AB sin α, FIG. 16 where Proj, AB and Proj, AB denote the projections of AB on the x-axis and the y-axis, respectively. 15. Applications of Projections. In mechanics and related subjects, forces and velocities are represented graphically by line segments. A force, say of 10 lb., is represented by a segment 10 units in length in the direction of the force. A velocity of 20 ft. per sec. is represented by a segment 20 units in length in the direction of motion. The projection upon a given line 7, of a segment representing a force, represents the effective force in the direction 7; this is called the component of the given force in the direction 7. Example 1. A weight of 50 lb. is placed upon a smooth plane inclined at an angle of 27° with the horizontal. What force acting directly up the incline will be required to keep the weight at rest? N R 30 sin27 27° M 127 Draw to some convenient scale a segment 50 units in length directly downward to represent the force exerted by the weight. Project this segment upon a line inclined at an angle of 27° with the horizontal. The length of this projection is 50 sin 27° = 22.7 nearly. This represents the component of the force down the plane. Therefore, a force of 22.7 lb. acting up the plane will be required to keep the weight at rest. EXERCISES VII. - PROJECTIONS FIG. 17 1. Find the horizontal and vertical projections of the segments: 2. A straight railroad crosses two north and south roadways a mile apart. The length of track between the roadways is 14 mi. A train travels this distance in 2 min. Find the components of the velocity of the train parallel to the roadways and perpendicular to them. Find the angle between the track and either roadway. 3. The eastward velocity of a certain train is 24 mi. per hour. The northward velocity is 32 mi. per hour. Find its actual velocity along the track and the angle the track makes with the east and west direction. 4. A car is drawn by means of a cable. If a force of 5000 lb. exerted along the track is required to pull the car, what force will be required when the cable makes an angle of 15° with the track? 5. Find the horizontal and vertical components of a force represented by a segment 30 units long at an angle of 40° with the horizontal. 16. The Use of Logarithms. Logarithms may be used to shorten computations involving multiplications, divisions, raising to powers or extracting roots, but not involving additions or subtractions. In much of the numerical work which follows, the use of logarithms is very advantageous in saving time and labor, but the student should bear in mind that logarithms are not necessary. They are merely convenient, and they belong no more to trigonometry than to arithmetic. One of the questions which a computer has to decide is whether or not it will be advantageous to use logarithms in a given problem. Tables usually contain (1) tables of logarithms; (2) tables of the trigonometric functions; (3) tables of logarithms of the trigonometric functions. The notation log tan 61° means the logarithm of the tangent of 61°; i.e. the tangent of 61° is a number, 1.804+, and the logarithm of this number 1.804+ is .25625. A formula which has been arranged so as to involve only products and quotients of powers and roots of quantities either known or easily computed is said to be adapted to logarithmic computation. Thus the formula h = √ a2 + b2, which gives the hypotenuse h of a right triangle in terms of the sides a and b, is not adapted to logarithmic computation. On the other hand, the formula b = √h2 — a2 = √(h + a)(h − a) which gives one side in terms of the hypotenuse and the other side is adapted to logarithmic computation because (h+a) and (h− a) are easily obtained from h and a. Thus, if the hypotenuse is 17.34 and one side is 12.27, the other side is Again, such an expression as 12.5 sin 42° 37', which might occur in solving a triangle, would be computed as follows: 17. Products with Negative Factors. To find by use of logarithms the product of several factors some of which are negative, the product of the same factors, all taken positively, is first obtained, and the sign is then determined in the usual way by counting the number of negative signs. Thus, to obtain the product of the four factors — 115, 23.41, — .6422, .1123, we write x = (115) (23.41)(.6422)(.1123); then and Here the fact that the first factor and the last two are negative is indicated by writing an (n) in parenthesis to the right of the corresponding logarithms. The product of the given factors all taken positively is 194.15; since the number of negative factors is odd the product is really — 194. 15. For other processes, see the Explanation of the Tables. EXERCISES VIII. RIGHT TRIANGLES - MISCELLANEOUS 1. Solve by means of logarithms the following right triangles, where h denotes the hypotenuse, other small letters the sides, and the corresponding capital letters the angles opposite those sides. 2. A tree stands on the opposite side of a small lake from an observer. At the edge of the lake the angle of elevation of the top of the tree is found to be 30° 58'. The observer then measures 100 ft. directly away from the tree and finds the angle of elevation to be 18° 26'. Find the height of the tree and the width of the lake. 3. From a point 250 ft. from the base of a tower and on a level with the base the angle of elevation of the top is 62° 32'. Find the height. 4. To determine the height of a tower, its shadow is measured and found to be 97.4 ft. long. A ten-foot pole is then held in vertical position and its shadow is found to be 5.5 ft. Find the height of the tower and the angle of elevation of the sun. 5. Find the length of a ladder required to reach the top of a building 50 ft. high from a point 20 ft. in front of the building. What angle would the ladder in this position make with the ground? |