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142 Sect 9.1– Applications Involving Right Triangles Objective 1: Finding the missing sides and angles of a right triangle. Given either two sides or a side and one of the acute angles of a right triangle, we can use our trigonometric ratios and the Pythagorean Theorem to solve for the missing sides and angles of a right triangle. Finding the missing sides and angles of the following. Round all sides to the nearest tenth and all angles to the nearest minute: E B Ex. 1 Ex. 2 18 ft 71.5˚ 23˚ C A F 13 in G Solution: Since ∠A and ∠B are complimentary, then m∠B = 90˚ – 23˚ = 67˚. We have the hypotenuse and the acute angle A. We can use the sine of 23˚ to find BC and the cosine of 23˚ to find AC. sin 23˚ = = Solution: Since ∠E and ∠G are complimentary, then m∠E = 90˚ – 71.5˚ = 18.5˚ = 18˚30'. We have the adjacent side and and the acute angle G. We can use the tangent of 71.5˚ to find the EF and the cosine of 71.5˚ to find EG. tan 71.5˚ = = To solve, multiply by 18: BC = 18•sin 23˚ = 18•0.3907… = 7.03… ≈ 7.0 ft cos 23˚ = = To solve, multiply by 13: EF = 13•tan 71.5˚ = 13•2.98… = 38.85… ≈ 38.9 in cos 71.5˚ = = To solve for BC, multiply by 18: AC = 18•cos 23˚ = 18•0.9205… = 16.56… ≈ 16.6 ft Thus, BC ≈ 7.0 ft. AC ≈ 16.6 ft and m∠B = 67˚. To solve for EG, multiply by EG and then divide by cos 71.5˚: EG•cos 71.5˚ = 13 EG = 13/cos 71.5˚ =13/0.317… = 40.97… ≈ 41.0 in Hence, EF ≈ 38.9 in, EG ≈ 41.0 in, and m∠E = 18˚30'. 143 Ex. 3 C 7 cm B Ex. 4 F 10.1 m 5 cm E A 15 m G Solution: We have the two legs of a right triangle. We can use the inverse tangent function Solution: We have the leg and the hypotenuse of a right triangle. we can use the inverse cosine to to find ∠B: tan B = to find ∠G: cos G = Thus, B = tan – 1( = ) = 35.53… ˚ Convert into DMS and round: 35˚32'15"6 ≈ 35˚32' Thus, m∠B = 35˚32' Since ∠A and ∠B are complimentary, then m∠A = 90˚ – 35˚32' = 54˚28' To find the length AB, we can use the Pythagorean Theorem: (AB)2 = (7)2 + (5)2 (AB)2 = 49 + 25 (AB)2 = 74 AB = = 8.60… ≈ 8.6 cm Thus, m∠B ≈ 35˚32', m∠A ≈ 54˚28', and AB ≈ 8.6 cm. Objective 2: Hence, G = cos – 1( = ) = 47.67… ˚ Convert into DMS and round: 47˚40'30"5 ≈ 47˚41' Therefore, m∠G = 47˚41' Since ∠E and ∠G are complimentary, then m∠E = 90˚ – 47˚41' = 42˚19' To find the length EF, we can use the Pythagorean Theorem: (10.1)2 + (EF)2 = (15)2 102.01 + (EF)2 = 225 – 102.01 = – 102.01 2 (EF) = 122.99 EF = =11.09… ≈ 11.1 m Hence, m∠G ≈ 47˚41', m∠E = 42˚19', and EF ≈ 11.1 m. Solving applications involving right triangles. In solving right triangles, we will use the fact that the two acute angles are complementary and we will use the Pythagorean Theorem. Right Triangle Theorem Let A, B, and C be the angles of the triangle and a, b, and c be the lengths of the corresponding sides. If C is a right angle, then A + B = 90˚ and c2 = a2 + b2 144 Solve the following: Ex. 5 A road rises 223 ft over a horizontal distance of 2500 ft. What is the angle of elevation? (round to the nearest tenth). Solution: First, draw a diagram: 223 ft θ 2500 ft We have the opposite and adjacent sides of a right triangle, so we want to use the inverse tangent function to find the angle opp 223 tan θ = = (use the inverse tangent) adj θ = tan – 1( Ex. 6 2500 223 ) 2500 = 5.097… ≈ 5.1˚ €For a € wheel chair accessible ramp to be ADA compliant, the ratio of the rise to the run of the ramp must be 1 to (12 or greater). A 29 ft 9 € in long ramp leading into a building climbs 2 ft 5 in. a) Is this ramp ADA compliant? b) If a new city ordinance states that the incline of a handicapped access can be no more than 4.5˚, does this ramp comply with the new ordinance? If not, how long does the ramp need to be? Round to the nearest inch. c) Another requirement for a ramp to ADA compliant is that is one 5 ft by 5 ft rest platform is required after a run of a maximum of 30 feet in a ramp. Does the new ramp require a rest platform? Solution: a) We will begin by drawing a picture: Length Rise ← Angle of Elevation Run The Length of the ramp corresponds to the hypotenuse, and the Rise and Run correspond to the two legs. Thus, we will need convert the measurements into inches and use the Pythagorean theorem to calculate the run of the ramp: 145 Rise = € ( ) + 5 in = 29 in Length = 29 ft 1 ( ) + 9 in = 357 in (Run)2 = (Length)2 – (Rise)2 = (357)2 – (29)2 = 127449 – 841 = 126608 Run = 126608 = 355.52… in The ratio of the Rise to the Run is € 29 to 355.52… Dividing by 29, we get: 1 to 12.269… Since 12.269… is greater than 12, the ramp is ADA compliant. b) Since the ramp climbs 29 in, then 357 in the height of the ramp is 29 in. This 29 in x will correspond to the opposite side of the incline. The length of the ramp is 357 in. This will correspond to the hypotenuse. This means we will need to use the inverse sine function to find the angle. 29 sin x = = (use the inverse sine) 357 x = sin – 1 ( 29 ) 357 = 4.659…˚ which is greater than 4.5˚ Thus, the current ramp does not comply with the ordinance. € climbs 29 in, then Since the ramp y the height of the ramp is 29 in. This 29 in 4.5˚ € will correspond to the opposite side of the incline. The length of the ramp is length of the ramp which is the hypotenuse. This means we will need to use the sine function. sin 4.5˚ = = y sin 4.5˚ = 29 29 y= = sin 4.5 29 y 29 0.07845... (multiply by y) (divide by sin 4.5˚) = 369.619… ≈ 370 in € But, 370 ÷ 12 = 30 with a remainder 10, so 370 in = 30 ft 10 in The ramp should be 30 ft 10 in to comply with the ordinance. € € c) To determine if the new ramp needs a rest platform, we will need to calculate the Run: (Run)2 = (Length)2 – (Rise)2 = (370)2 – (29)2 = 136900 – 841 = 136059 Run = 136059 = 368.86… in Converting to feet, we get: 368.86…/12 = 30.738… feet which is greater than 30 feet so the new ramp will need a rest platform. € 146 In navigation, the bearing or direction is measured as the acute angle either east or west of north or south. Thus, if a ship is traveling 35˚ east of north, it's bearing would be N35˚E while a airplane flying 22˚ west of south would have a bearing of S22˚W. Identify the bearing for each point: Ex. 7 N A B 77˚ 38˚ W E C 65˚ D 5˚ S Solution: A) The bearing is N38˚E. B) The bearing is N77˚W. C) The bearing is S65˚E D) The bearing is S5˚W. Solve the following: Ex. 8 A ship leaves port in the Florida Keys with a bearing of S85˚E with a speed of 20 knots. After two hours, the ship turns 90˚ toward the north. After 3 additional hours of maintaining the same speed, what is the bearing of the ship from the port? Solution: We begin by drawing a diagram of the situation. The ship travels 20(2) or 40 nautical miles S85˚E from the Florida Keys. it then turns 90˚ toward the north and travels 20(3) = 60 nautical miles. If we connect a line from the Florida Keys to the ending point, we will form a right triangle. We will label the angle formed by the 40 nautical θ 40 85˚ 60 147 mile side and the line connecting the Florida Keys to the ending point θ. Since we have the opposite side and the adjacent side, we will use the tangent function to find θ. tan(θ) = opp adj = 60 40 = 1.5. Thus, θ = tan – 1(1.5) = 56.3099…˚ ≈ 56.3˚ Since the 40 nautical mile side had a bearing of S85˚E, the angle between that side and the positive x-axis is 90˚ – 85˚ = 5˚. € €The angle between the positive x-axis and the line connecting the Florida Keys to the ending point is θ – 5˚ = 56.3˚ – 5˚ = 51.3˚. Finally, the angle between the line connecting the Florida Keys to the ending point and the positive y-axis is 90˚ – 51.3˚ = 38.7˚ This means the bearing from the Florida Keys to the ship is N38.7˚E.