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Jeff Bivin -- LZHS Right Triangle Trigonometry By: Jeffrey Bivin Lake Zurich High School Last Updated: December 1, 2010

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Jeff Bivin -- LZHS SOH CAH TOA θ opposite hypotenuse adjacent

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Jeff Bivin -- LZHS Reciprocal Identities θ opposite hypotenuse adjacent

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Jeff Bivin -- LZHS Reciprocal Identities θ opposite hypotenuse adjacent

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Jeff Bivin -- LZHS Reciprocal Identities θ opposite hypotenuse adjacent

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Jeff Bivin -- LZHS a Find the sides. 1 A C B b c 2

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Jeff Bivin -- LZHS a Find the sides. 10 A C B b c

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Jeff Bivin -- LZHS a Find the sides. 17 A C B b c

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Jeff Bivin -- LZHS a Find the sides. A C B b c 9

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Jeff Bivin -- LZHS Find the sides. 22 o 68 o c A C B b 15 a

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Jeff Bivin -- LZHS Find the sides. 34 o 56 o c A C B b a25

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Jeff Bivin -- LZHS Find the angles and the 3 rd side. θ β

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Jeff Bivin -- LZHS Find the angles and the 3 rd side. θ 6 11 β

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Jeff Bivin -- LZHS

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Quotient Identities θ opposite hypotenuse adjacent

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Jeff Bivin -- LZHS Pythagorean Identities θ opposite hypotenuse adjacent

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Jeff Bivin -- LZHS Using Pythagorean Identities Find cosθ if sinθ =

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Jeff Bivin -- LZHS Using Pythagorean Identities Find secθ if tanθ =

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Jeff Bivin -- LZHS Using Pythagorean Identities Find sinθ if cotθ =

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Jeff Bivin -- LZHS θcos(θ)sin(θ) 10 o o o o o o o o o o Co-function Identities Use your calculators to evaluate each of the following. Complimentary Angles cos(θ) = sin(90 o – θ) and sin(θ) = cos(90 o – θ)

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Jeff Bivin -- LZHS Co-function Identities Complimentary Angles cos(θ) = sin(90 o – θ) and sin(θ) = cos(90 o – θ) sec(θ) = csc(90 o – θ) and csc(θ) = sec(90 o – θ)

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Jeff Bivin -- LZHS Co-function Identities Complimentary Angles cos(θ) = sin(90 o – θ) and sin(θ) = cos(90 o – θ) sec(θ) = csc(90 o – θ) and csc(θ) = sec(90 o – θ) tan(θ) = cot(90 o – θ) and cot(θ) = tan(90 o – θ)

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Jeff Bivin -- LZHS Co-function Identities Complimentary Angles cos(θ) = sin(90 o – θ) and sin(θ) = cos(90 o – θ) sec(θ) = csc(90 o – θ) and csc(θ) = sec(90 o – θ) tan(θ) = cot(90 o – θ) and cot(θ) = tan(90 o – θ) θ 90 o - θ a b c

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Jeff Bivin -- LZHS (a, b) (a, -b) (b, a) t -t t 90 o -t

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Jeff Bivin -- LZHS

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A surveyor is standing 45 feet from the base of a large tree. The surveyor measures the angle of elevation from the ground to the top of the tree to be 67.5 o. Find the height of the tree. 45 feet 67.5 o h

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Jeff Bivin -- LZHS An airplane flying at 4500 feet is on a flight path directly toward an observer. If 30 o is the angle of elevation from the observer to the plane, find the distance from the observer to the plane feet 30 o d

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Jeff Bivin -- LZHS In traveling across flat land a driver noticed a mountain directly in front of the car. The angle of elevation to the peek is 4 o. After the driver traveled 10 miles, the angle of elevation was 11 o. Approximate the height of the mountain. 10 mi h 11 o x 10 + x 4o4o

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Jeff Bivin -- LZHS A flagpole at the top of a tall building (and at the edge of the building) is know to be 45 feet tall. If a man standing down the street from the building calculates the angle of elevation to the top of the building to be 55 o and the angle of elevation to the top of the flagpole to be 57 o. Find the height of the building. 55 o h o d

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Jeff Bivin -- LZHS An observer standing on the cliff adjacent to the ocean looks out and sees an airplane flying directly over a ship. The observer calculates the angle of elevation to the plane to be 14 o and the angle of depression to the ship to be 27 o. How high above the ship is the airplane if we know that the ship is 1.5 miles from shore? 14 o b 27 o p Distance of plane above ship = p + b = 1.5 mi

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Jeff Bivin -- LZHS In Washington, D.C., the Washington Monument is situated between the Capitol and the Lincoln Memorial. A tourist standing at the Lincoln Memorial tilts her head at an angle of 7.491° in order to look up to the top of the Washington Monument. At the same time, another tourist standing at the Capitol steps tilts his head at a 5.463° to also look at the top of the Washington Monument. Find the distance from the Lincoln Memorial to the Washington Monument.

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