1. CHAPTER EIGHT Alec Rodriguez Jack Wells Chris “the Bottman” Bott

2. 8.1 Similarity in Right Triangles  Theorem 8-1 Right Triangle Similarity  If an altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. C A B D

3. Geometric Mean  The mean between two numbers in a geometric sequence.  2,4,8,16,32  a/x =x/b  Ex. 2/x = x/32  Answer: 8

4. Corollary 1  When the altitude is drawn to the hypotenuse of a right triangle, the length of the altitude is the geometric mean between the segments of the hypotenuse. C A B D AD/CD = CD/BD

5. Corollary 2  When the altitude is drawn to the hypotenuse of a right triangle, each leg is the geometric mean between the hypotenuse and the segment of the hypotenuse that is adjacent to that leg. C A B D AB/AC= AC/AD AB/BC = BC/BD

6. Challenge Find AB, AC, CD, CB 9 16 X Y Z Triangle ABC is a right triangle.

7. The Pythagorean Theorem  In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.  A 2 + B 2 = C 2  Pythagoras.

8. Challenge 6 8 X 1. Find X.2. Find C. 45° 90° 2√2 C

9. 8-3 Converse of the Pythagorean Theorem B A C C 2 = A 2 + B 2 Right Triangle C 2 < A 2 + B 2 Acute Triangle C 2 > A 2 + B 2 Obtuse Triangle

10. 8-3 Converse of the Pythagorean Theorem Example: Is the triangle acute, obtuse, or right? 13 2 + 15 2 ____ 29 2 169 + 225 ____ 841 394 < 841 The triangle is acute. 15 13 29 12 2 + 18 2 ____ 19 2 144 + 324 ____ 361 468 > 361 The triangle is obtuse. 18 12 19

11. 45 Special Right Triangles 1. 45 – 45 – 90 General Rule a a a 2 45

12. More Special Right Triangles 2. 30 – 60 – 90 General Rule a a 3 2a 60 30

13. Even More Special Right Triangles Challenge 7 45 X Find X. 30 4 3 Y Find Y.

14. Sine  Formula : sin Ѳ =Opposite Hypotenuse Ѳ Adjacent Opposite Solve for x: – Sin20=4/x – Multiply each side by x – X(sin20)=4 – Divide each side by sin20 – X=11.695 4 20 ⁰ x

15. Cosine  Formula : cos Ѳ =Adjacent Hypotenuse  Solve for x:  Cos67=x/120  Multiply both sides by 120  120(cos67)=x  Multiply 120 and cos67  46.88=x Ѳ Hypotenuse Adjacent Opposite x 120 67 ⁰

16. Tangent  Formula : tan Ѳ =Opposite Adjacent Ѳ Hypotenuse Adjacent Opposite Solve for x Tan42=x/5 Multiply each side by 5 5(tan42)=x Multiply 5 and tan42 4.5=x 5 x 42 ⁰

17. SOH-CAH-TOA  An easy way to remember all of these formulas is by using SOH CAH TOA  SOH - (sine) opposite over hypotenuse  CAH - (cosine) adjacent over hypotenuse  TOA - (tangent) opposite over adjacent

18. Applications of Right Triangle Trigonometry  How to solve: 1. Tan2 ⁰ = 25/x 2. x = 25/tan2 ⁰ 3. x = 716.3 Angle of elevation Angle of depression horizontal Line of sight 2⁰2⁰ 2⁰2⁰ 25 x

19. Exercises  Solve for x and y: x y 35 37 ⁰ When the sun’s angle of elevation is 57 ⁰, a building casts a shadow 21m long. How high is the building? 21m 57 ⁰

20. Last Exercise  An observer is located 3km from a rocket launch site sees a rocket at an angle of elevation of 38 ⁰. How high is the rocket at that moment?