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Warm up A.14.1 B.17.4 C.19.4 D.21.3 Find x. Round to the nearest tenth.
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Warm up A.9.5 B.15.9 C.23.7 D.30.8 Find x. Round to the nearest tenth.
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Unit 6 Lesson 3 Inverse Trig Real World Application I can use trigonometric ratios to find angle measures in right triangles. I can use trigonometric ratios to solve word problems.
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Inverse Trigonometric Ratios
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Remember… Solving Trigonometric Equations There are only three possibilities for the placement of the variable ‘x”. 5 Sin = Sin = 0.48 X = Sin (0.48) X = 28.6854 Sin 25 = x = (12) (0.4226) x = 5.04 cm 0.4226 = Sin 25 = 0.4226 = x = x = 28.4 cm
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Use a calculator to find the measure of P to the nearest tenth. The measures given are those of the leg adjacent to P and the hypotenuse, so write the equation using the cosine ratio. KEYSTROKES: [COS] 13 1946.82644889 2nd( ÷)ENTER Answer: So, the measure of P is approximately 46.8°.
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A.44.1° B.48.3° C.55.4° D.57.2° Use a calculator to find the measure of D to the nearest tenth.
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A.34.7 B.43.8 C.46.2 D.52.5 Find x. Round to the nearest tenth.
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Solve the right triangle. What are we looking for? ALL sides of the triangle ALL angles of the triangle Round side measures to the nearest hundredth and angle measures to the nearest degree.
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Step 1Find m A by using a tangent ratio. 29.7448813≈m AUse a calculator. So, the measure of A is about 30 . Definition of inverse tangent
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Step 2Find m B using complementary angles. m B≈60Subtract 30 from each side. So, the measure of B is about 60 . 30 + m B≈90m A ≈ 30 m A + m B=90Definition of complementary angles
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Step 3Find AB by using the Pythagorean Theorem. (AC) 2 + (BC) 2 =(AB) 2 Pythagorean Theorem 7 2 + 4 2 =(AB) 2 Substitution 65=(AB) 2 Simplify. Take the positive square root of each side. 8.06≈ ABUse a calculator. Answer: m A ≈ 30, m B ≈ 60, AB ≈ 8.06
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A.m A = 36°, m B = 54°, AB = 13.6 B.m A = 54°, m B = 36°, AB = 13.6 C.m A = 36°, m B = 54°, AB = 16.3 D.m A = 54°, m B = 36°, AB = 16.3 Solve the right triangle. Round side measures to the nearest tenth and angle measures to the nearest degree.
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Angle of elevation The angle formed by a horizontal line (usually the ground) and an observer’s line of sight to an object above
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Angle of Depression The angle formed by a horizontal line (usually imagined) and an observer’s line of sight to an object below
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Angle of Elevation CIRCUS ACTS At the circus, a person in the audience at ground level watches the high-wire routine. A 5-foot-6- inch tall acrobat is standing on a platform that is 25 feet off the ground. How far is the audience member from the base of the platform, if the angle of elevation from the audience member’s line of sight to the top of the acrobat is 27°? Make a drawing.
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Since QR is 25 feet and RS is 5 feet 6 inches or 5.5 feet, QS is 30.5 feet. Let x represent PQ. Multiply both sides by x. Divide both sides by tan Simplify.
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A.37° B.35° C.40° D.50° DIVING At a diving competition, a 6-foot-tall diver stands atop the 32-foot platform. The front edge of the platform projects 5 feet beyond the ends of the pool. The pool itself is 50 feet in length. A camera is set up at the opposite end of the pool even with the pool’s edge. If the camera is angled so that its line of sight extends to the top of the diver’s head, what is the camera’s angle of elevation to the nearest degree?
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DISTANCE Maria is at the top of a cliff and sees a seal in the water. If the cliff is 40 feet above the water and the angle of depression is 52°, what is the horizontal distance from the seal to the cliff, to the nearest foot? Make a sketch of the situation. Since are parallel, m BAC = m ACD by the Alternate Interior Angles Theorem.
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Let x represent the horizontal distance from the seal to the cliff, DC. c = 52°; AD = 40, and DC = x Multiply each side by x. Divide each side by tan 52°. The seal is about 31 feet from the cliff.
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A.19 ft B.20 ft C.44 ft D.58 ft Luisa is in a hot air balloon 30 feet above the ground. She sees the landing spot at an angle of depression of 34 . What is the horizontal distance between the hot air balloon and the landing spot to the nearest foot?
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DISTANCE Vernon is on the top deck of a cruise ship and observes two dolphins following each other directly away from the ship in a straight line. Vernon’s position is 154 meters above sea level, and the angles of depression to the two dolphins are 35° and 36°. Find the distance between the two dolphins to the nearest meter.
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UnderstandΔMLK and ΔMLJ are right triangles. The distance between the dolphins is JK or JL – KL. Use the right triangles to find these two lengths. PlanBecause are horizontal lines, they are parallel. Thus, and because they are alternate interior angles. This means that
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Multiply each side by JL. Divide each side by tan Use a calculator. Solve
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Answer: The distance between the dolphins is JK – KL. JL – KL ≈ 219.93 – 211.96, or about 8 meters. Multiply each side by KL. Use a calculator. Divide each side by tan
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A.14 ft B.24 ft C.37 ft D.49 ft Madison looks out her second-floor window, which is 15 feet above the ground. She observes two parked cars. One car is parked along the curb directly in front of her window and the other car is parked directly across the street from the first car. The angles of depression of Madison’s line of sight to the cars are 17° and 31°. Find the distance between the two cars to the nearest foot.
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