9-1 The Tangent Ratio 4/26/17 T an A = pposite djacent O A Objective: To use tangent ratios to determine side lengths in right triangles. B B T an A = pposite djacent O Hypotenuse A Opposite A A C C Adjacent What is Soh Cah Toa? Soh Cah TBD…

Reciprocals! T 5 4 V U 3 Ex: Find tan T = opp = UV = 3 adj TV 4 Find tan U = opp = TV = 4 adj UV 3 U 5 3 T 4 V Write tangent ratios for K and J. How is tan K related to tan J? J 3 L 7 K T 5 4 V U 3 Tan K = 3 7 Tan J = 7 3 Reciprocals!

Make sure the calc is in DEGREE mode!!!! Find x Ex: Use a calculator 86o tan 86o = x 50 50 50(tan 86o) = x In calc: 50 • tan 86 = 715.03331 About 715 Find w to the nearest tenth. 1) 10 2) y 54o 28o w 10 3) 2.5 w 57o 33o x x Tan 28 = 10 y Tan 54 = w 10 y •Tan 28 = 10 10 •(Tan 54) = w y = 10 e = 18.8 w = 13.8 (Tan 28) Tan 57 = w 2.5 Tan 33 = 2.5 w w = 3.8 w = 3.8 It works either way!

Ex: Find m x to the nearest degree. tan x = 6 = .75 H 8 Tan-1 (.75) is asking “what angle has a tangent of .75?” It is called the Inverse Tangent. Tan-1  that is NOT a power! Ex: Find m x to the nearest degree. tan x = 6 = .75 H 8 m x = tan-1(0.75) 6 10 tan-1 .75 = 36.869898 m x 37 B 8 X Find m Y to the nearest degree. P 100 T 41 Y m y = Tan-1 (100/41) = 68° You might want to do 100/41 first, then calculate the Tan-1 depending on how your calc works.

x K 8 L 15 M Write the tangent ratio for K Write the tangent ratio for M Find m M to the nearest degree Find x to the nearest whole number 1) 62 2) 90 40o x 72o 15 8 8 15 m M = Tan-1 (8/15) = Tan-1 (.5333) = 28° Tan 40 = x 62 x x Tan 72 = 90 x x = 52 x = 29

(When the circle has a radius of 1, that is.) Assignment: Page 472 #1 – 20