2. Refer to the figure of Example 2. Suppose that CB is 85 ft. and angle B is 40°. Find AB. 3. In the same figure, if CB is 60 ft., and angle B is 40°, find AC. 4. Let A and B be two points on opposite sides of a house. Suppose they can both be seen from point C, and suppose that AC, BC, and angle C have the measures marked on the figure. 100 ft. 60° 80 ft. B Draw this figure to scale. Determine the approximate distance from A to B. 5. Take two points on the blackboard. Suppose that you want the distance between them but are not allowed to measure that distance directly. Take measurements from a third point C, as in Example 4. Then make a scale drawing of the figure and find AB. 6. (Outdoor Exercise.) Find the height of some telephone pole or building as is done in Example 2 on page 222. 7. Find the distance from a point on the window sill of your room to some tree which can be seen from it, by making measurements like those in Example 1. 25° 35° C 100ft. D B 8. Make a scale drawing of the figure at the right, using the measurements marked in the figure. Determine the approximate length of AB. (This is an indirect way of measuring the height of an object when it is impossible to measure up to the foot of the perpendicular distance from its top to the ground.) 126. Indirect measurements by means of a graph. The following graph reviews what you learned on that subject in Book I. 1. How many pounds of round steak would $1 purchase in 1907? in 1916? 2. In what two years was it possible to purchase more for $1 than during the preceding year? 3. In general, has the amount which could be purchased decreased or increased? How do you know? 4. Determine the number of pounds which you can buy for $1 in your village or city. Draw upon the figure, in pencil, a segment representing this number of pounds. 127. Graphing negative numbers. The curved black line in the figure on page 225 is the graph of the temperature readings at certain hours of a certain day in a certain place. Notice the heavy black horizontal line. It is called the horizontal axis. On it, the hours 6, 7, 8, etc., are marked at equal distances. Three spaces horizontally are allowed for each hour, or one space for every 20 minutes. Notice the heavy black vertical line. It is called the vertical axis. On it, the temperature readings above zero are marked +, and those below zero are marked Five spaces vertically are allowed for five degrees, or one space for one degree. This graph was drawn from temperature readings at every hour from 6 A.M. to 8 P.M. You can tell what the temperature was at various times during the day. At 11 A.M. the vertical distance from the horizontal axis to the temperature curve is 7 spaces upward. Therefore the temperature was +7° or 7° above zero. At 6 A.M. the distance from the horizontal axis to the temperature curve is 8 spaces downward. Hence the temperature was -8° or 8° below zero. EXERCISE 106 What was the temperature at: 1. 7 A.M. 3. 9 A.M. 5. 12 M. 7. 5 P.M. 9. 9:40 A.M. 2. 8 A.M. 4. 10 A.M. 6. 2 P.M. 8. 8 P.M. 10. 5:30 P.M. 11. At what time in the morning was the temperature zero? 12. Obtain the temperature readings in your village or city for yesterday. Draw a graph representing them. 128. Graphing the relation between two numbers connected by a simple formula. Example. 1. If the altitude of a rectangle is 5.5 ft. and its base is b ft., its area A is given by the formula Draw a graph representing this relation. A=5.56 sq. ft. Solution. - 1. The graph of this relation is obtained as follows: The graph is the oblique line in the figure below. The scale for b is marked on the vertical axis; 5 spaces represent 2 ft., or 1 space or .4 ft. The scale for A is marked on the horizontal axis; 10 spaces represent 10 sq. ft., or 1 space 1 sq. ft. When b is 2, A is 11. This is represented by point R, obtained by going 2 units up and 11 units to the right. S represents the fact that A is 22 when b is 4. Similarly for points T, X, and Y. The oblique line is drawn through the points. It happens to be a straight line. 2. This graph enables us to solve two problems. a.- - Find the area of a rectangle of altitude 5.5 ft. for a given base. Thus, let b=2.8, find A. Begin at W on the vertical axis, which represents b=2.8. Follow the dotted line to Z on the graph, and then to K on the horizontal axis. K is at the point where A = 15.5. .. Aabout 15.5, when b=2.8. Check. 2.8 5.5 140 140 15.40 b. - Find the base of the rectangle of altitude 5.5 ft. when the area is given. Thus, find the base when the area is 35 sq. ft. Begin at 35 on the A axis. Follow the dotted line to M on the graph, and then to N on the vertical axis. N is at the point where b = 6.4. In both problems the results are approximately correct. That is all that can be expected from such solutions. Nevertheless in many mathematical problems that is not only sufficient, but is the best that can be done. EXERCISE 107 From the graph above: 1. Find A when b = 8.4. 2. Find A when b=6.8. 3. Find A when b=5.2. 4. Find A when b=3.6. 5. Find b when A=40. 6. Find b when A=46. 7. Find b when A = 25. 8. Find b when A=30. |