This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1920 Excerpt: ...EF because the sides are not similar or are not in the same order on each side of the equality sign. In practical problems it is customary to draw the two triangles together (Fig. 53). Thus, triangle ADE is similar to triangle ABC and we may form such proportions as AD: AC:: AE: AB AB: BC:: AE: DE AE: AB:: DE: BC, etc. ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1920 Excerpt: ...EF because the sides are not similar or are not in the same order on each side of the equality sign. In practical problems it is customary to draw the two triangles together (Fig. 53). Thus, triangle ADE is similar to triangle ABC and we may form such proportions as AD: AC:: AE: AB AB: BC:: AE: DE AE: AB:: DE: BC, etc. Example. This principle may be applied to a practical problem in finding the height of towers, chimneys and similar objects. In Fig. 54, H, the height of the tower is the distance to be determined. Measure on the ground a level distance of 150 feet from the center of the base of the tower. From the point A, sight towards the top of the tower through the top of a stick set upright in the ground. Then measure the distance from A to E, (the bottom of T k20'j' Fig. 54. Measuring a Tower. the stick). We will find in this case that AE is twenty feet. Then measure the height of the upright stick ED, which we will find in this instance to be 10 feet. Referring to Fig. 53 we find that our measuring has followed the lines of two similar right triangles of which certain dimensions are known. For instance the side AB equals 150 feet, the side EA equals 20 feet, the side ED equals 10 feet. The corresponding angles are all equal (see Fig. 53). We now have the necessary terms to form a proportion and by this means can find H, the height of the tower. Solution. H: 150:: 10: 20. Referring to the three principles on page 68 we find that 20 H = 10 X 150, or H = 10 15-5 15 By cancellation, II = = 75 feet. '20 10 Example. Another interesting application of the comparison of different heights as determined by ratio and proportion may be found by comparing the heights of objects with their respective shadows. Suppose, for example, that at a given time of the day a
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