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6. SOLVING RIGHT TRIANGLES In the right triangle ACB shown in Figure 6.1, the angles are denoted by α at vertex A, β at vertex B, and γ at vertex C. The lengths of the sides opposite the angles α , β , and γ are denoted by a, b, and c. Note that angles α and β are complementary acute angles, that angle γ is a right angle, and that c is the hypotenuse of right triangle ACB. Therefore, a sin α = cos β = c b cos α = sin β = c a tan α = cot β = b • BB c csc α = sec β = a c sec α = csc β = b b cot α = tan β = a β c a γ α • • b AA CC Figure 6.1 If the lengths of two sides of a right triangle are given, or if one side and an acute angle are given, then these formulas san be used to solve for the remaining angles and sides of the triangle. The procedure is called solving the right triangle. Example 6.1 ---------------------------- -----------------------------------------------------------Solve the right triangle ACB labeled as in Figure 6.2 if β = 30 o and a = 24. We must find b, c¸ and α . Because α and β are complementary, •B α = 90 o − β = 90 o − 30 o = 60 o . c 30 Figure 6.2 o 24 γ α • A To find b, we use the fact that tan β b • Therefore, b = a tan β C = b , a so b = a tan β . 3 = 24 tan 30o = 24 = 8 3 . To find c, we use the fact that sec β 3 = c , so c = a sec β . a Therefore, c = a sec β 2 3 = 16 3 = 24 sec 30 o = 24 Of course, we could have instead used the Pythagorean theorem to find c = 3 . a2 + b2 . _______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ Unless the special angles 30 o , 45 o , or 60 o are involved, it is necessary to use a calculator or a table of trigonometric functions to solve a right triangle. You should always keep in 40 mind that the solutions obtained in these ways are approximations. In this section, whenever such approximations are involved, we will round off all angles to the nearest hundredth of a degree, and all side lengths to four significant digits. Example 6.2 ---------------------------- -----------------------------------------------------------In each case, assume that right triangle ACB is labeled as in Figure 6.1. Then solve the triangle and sketch it. ( i ) b = 31, α = 43.33o . We must find a, c, and β . Because α and β are complementary, β = 90 o − α = 90 o − 43.33o = 46.67 o . a To find a, we notice that tan α = , so a = b tanα . b Using a calculator, we find that tan α = tan 43.33o = 0.943341386. Therefore, rounded off to four significant digits, a = b tan α = 31 tan 43.33o = 29.24 . b b , so c = . c cos α Therefore, rounded off to four significant digits, b 31 31 so c = = = = 42.62 o cos α 0.727413564 cos 43.33 To find c, we notice that cos α •B β = 46.67 o c=42.62 a=29.24 = γ α = 43.33o • • b=31 A C Figure 6.3 . Once again, we could have instead used the Pythagorean theorem to solve for c: c = a2 + b2 = (29.24)2 + 312 = 42.61. The discrepancy in the last decimal place was caused by using the rounded off value for a. The result c = 42.62 is actually more accurate. The triangle is shown in figure 6.3. ( ii ) a = 4, b = 3 . Figure 6.4 •B By the Pythagorean theorem: β = 36.87 o a2 + b2 = 16 + 9 = 5. a 4 Now we have sin α = = = 0.8 . c 5 c = c=5 α = arcsin 0.8 = 53.13 o . It follows that a=4 β = 90 o −α = 90 o − 53.13 o = 38.87 o . The triangle is shown in Figure 5.4. • A α = 53.13o b=3 γ • C _______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________ 41 Example 6.3 ---------------------------- -----------------------------------------------------------Consider a church with a steeple, as shown in the Figure 6.5. The problem is to calculate the height of the steeple while standing on the ground. To find the height, mark a point B on the ground directly below the steeple and another point A that is 200 feet away on the ground. At A , measure the angles of elevation α and β to the top and bottom of the steeple. This is all the information you need. Figure 6.5 Let x be the height of the steeple and y be the distance from the ground to the bottom of the steeple. Suppose that α = 35 o and β = 26 o . x+ y y Then tan 35 = and tan 26 o = . 200 200 Solving for y in the second equation and substituting into the first yields: • o tan 35 o = x • x + 200 tan 26 o . 200 y α = 35o So x = 200 tan 35 o - 200 tan 26 o =200 (.7002-.4877) = 42.5 feet. • β = 26 o • 200 ft A B Section 6 Problems--------------- ------- ----------------------------------------------------------In problems 1-3, assume that the right triangle is labeled as in the figure shown. In each case, solve the triangle and sketch it. When approximations are involved, round off angles to the nearest hundredth of a degree and side lengths to four significant digits. •B β c a γ α • • 1. a = 10, c =10 2 2. c =100, β = 41o 3. b = 91, β = 30 o A 42 b C 4. Find the area of the shaded region of the right triangle ABC shown in the figure. Round your answer to four significant digits. 5. The house in the figure is 24 ft. wide. The ridge is 9 ft higher than the side walls, and the rafters project 1.5 ft beyond the sides of the house. How long are the rafters? 6. A belt is to go around two pulleys of radii 1 inch and 2 inches with centers 6 inches apart. How long should the belt be? Hint: To find the length of the upper half of the belt begin by computing: (i) The distance, d, with similar triangles. (ii) The length, h, using the Pythagorean Theorem. (iii) The angles θ, α and β using right triangle trigonometry. 43 7. From the top of a lighthouse 120 feet above sea level, the angle of depression (the downward angle from the horizontal) to a boat adrift ion the sea is 9.4 o . How far from the foot of the lighthouse is the boat? 8. From a window in an office building, I am looking at a television tower that is 600 meters away (horizontally). The angle of elevation of the top of the tower is 19.6 o and the angle of depression of the base of the tower is 21.3o . How tall is the tower? 9. Find x in the figure. 10. The vertical distance from the first to the second floor of a certain department store is 28 feet. The escalator, which has a horizontal reach of 86 feet, takes 25 seconds to carry a person between floors. How fast does the escalator travel? 11. Consider two circles both of radius r and with the center of one lying on the rim of the other. Find the area of the common part of the two circles. 12. The Great Pyramid is about 480 feet high and its square base measures 760 feet on a side. Find the angle of elevation, β , of one of its edges. 44 13. Find the angle between a principal diagonal and a face of a cube. 14. A regular hexagon (6 equal sides) is inscribed in a circle of radius 4. Find the perimeter, P , and the area, A, of this hexagon. 15. Find the area of the regular 6-pointed Star of David (formed by two equilateral triangles) that is inscribed in a circle of radius 1. 16. Find the area of the regular 5-pointed star (the pentagram) that is inscribed in a circle of radius 1. 45