Aim: How do we use the principals of trigonometry to solve problems?

1. Aim: How do we use the principals of trigonometry to solve problems? Do Now:

2. Leg opposite 540 w w Leg adjacent to 540 10 10 Find the value of w to the nearest tenth. 10 10 represents the side adjacent to the 540 angle. 540 w w represents the side opposite the 540 angle. Tangent ratio Tan 540 = Tan 540 = 1.37638192 = w = (10)1.37638192 = 13.7638192 . . . w 13.8

3. Height h m. 520 520 h All but two of the pyramids built by the ancient Egyptians have faces inclined at 520. 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 520? 82 m 82 m Tan 520 = h/41 1.279941. . . = h/41 (41)1.279941. . . = h 52.4776. . . = h 52 meters h 41 m

4. 1st Degree Trig Equations algebraic trigonometric what values of x make this statement true? what values of  make this statement true? conditional equalities

5. QI QII QIII QIV 1st Degree Trig Equations cosine is positive in QI and QIV

6. Beware! no solution

7. QI QII QIII QIV Model Problem Find  for values between 0 and 360o to the nearest degree for 3 tan  - 5 = 7 tan is positive in QI and QIII 76o and 256o

8. QII QIV QI QIII 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

9. Model Problem Solve for : 5 (sin  + 3) = sin  + 12 for all values between 0 and 360o to the nearest minute. 5 (sin  + 3) = sin  + 12 sine is negative in QIII and QIV

10. QII QI QIII QIV Model Problem Solve for : 5 (sin  + 3) = sin  + 12 for all values between 0 and 360o to the nearest minute.

11. Regents Prep • If 0 < < 360o, find all values of  that satisfy the equation -4 cos  = 1 • 104o, 256o • 76o, 104o • 24o, 104o, 156o, 256o • 76o, 104o 156o, 256o