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Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:

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Presentation on theme: "Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:"— Presentation transcript:

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2 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Aim: How do we use the principals of trigonometry to solve problems? Do Now:

3 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Find the value of w to the nearest tenth. w 10 represents the side adjacent to the 54 0 angle. w represents the side opposite the 54 0 angle. Tangent ratio Tan 54 0 = Leg opposite 54 0 Leg adjacent to 54 0 Tan 54 0 = w 10 1.37638192 = w 10 w = (10)1.37638192 = 13.7638192... 10 54 0 w  13.8

4 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. All but two of the pyramids built by the ancient Egyptians have faces inclined at 52 0. Suppose an archaeologist discovers the ruins of a pyramid. Most of the pyramid has eroded, but she is able to determine that the length of a side of the square base is 82m. How tall was the pyramid, assuming its faces were inclined 52 0 ? 52 0 h Tan 52 0 = h/41 1.279941... = h/41 (41)1.279941... = h 52.4776... = h 52 meters  h 41 m 82 m 52 0 Height h m.

5 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. 1 st Degree Trig Equations algebraictrigonometric what values of x make this statement true? conditional equalities what values of  make this statement true?

6 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. 1 st Degree Trig Equations QIQIIQIIIQIV cosine is positive in QI and QIV

7 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Beware! no solution

8 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Model Problem Find  for values between 0 and 360 o to the nearest degree for 3 tan  - 5 = 7 tan is positive in QI and QIII 76 o and 256 o QIQIIQIIIQIV

9 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Model Problem Solve for cos  : cos  = 3 cos  + 1 for all values between 0 and 2 . cos  = 3 cos  + 1 cosine is negative in QII and QIII QI QII QIII QIV

10 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Model Problem Solve for  : 5 (sin  + 3) = sin  + 12 for all values between 0 and 360 o to the nearest minute. sine is negative in QIII and QIV 5 (sin  + 3) = sin  + 12

11 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Model Problem Solve for  : 5 (sin  + 3) = sin  + 12 for all values between 0 and 360 o to the nearest minute. QI QII QIIIQIV

12 Aim: Solving Linear Trig Equations Course: Alg. 2 & Trig. Regents Prep If 0 <  < 360 o, find all values of  that satisfy the equation -4 cos  = 1 1.104 o, 256 o 2.76 o, 104 o 3.24 o, 104 o, 156 o, 256 o 4.76 o, 104 o 156 o, 256 o


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