2. The Height Proportion Base of right Triangles

3. Imagine 2 similar right triangles 3m 4m 6m 8m Height Base = 3 4 = 6 8 =0.75 decimal The height of the larger triangle will be the key unknown…..

4. h 4.35m 7.61m 6.38m 50 O h 4.35 = 7.61 6.38 h 4.3 = 1.19 X 4.3 4.3 X h = 5.12m

5. h 7.24m 15.81m 26.31m 31 O h 7.24 = 15.81 26.31 h 7.24 = 0.60 X 7.24 7.24 X h = 4.35m

6. Height Base = H 4.1 = 1 1.7 tree meter stick 1.7H = 4.1 H = 2.4 m Let’s try some for real, as a group…outside… yes…….. KEY TRIANGLE

7. Imagine trying to find the height of a tree… 1m 1.7m 4.1m ? m 1.Take advantage of the fact that we can model these 2 situations with similar triangles 2. Create a proportion 3. Solve for an unknown that we can not physically obtain!!!

8. A more efficient way… Imagine, as the sun moves across the sky, that it creates many different angles for our triangles…

9. Technically, there is an infinite number of triangles that could be used in our proportions…. We need to be more efficient than that…. We will limit our triangles to under 100 000

10. The good news…. Instead of use having to go out and measure each possible triangle individually to get our KEY triangle….someone has done that for us…… Not only have they done that for us, they have also given cool names to 3 of the most useful ratios…. For example……

11. We need a better name for the height / base ratio… Since both sides involved are touching the right angle… height base The latin word “tangens” was used… Tangens was eventually converted to Tangent, or TAN.

12. O “theta” adjacent opposite hypotenuse Consider the ratio:

13. This ratio was first studied by Hipparchus (Greek), in 140 BC.

14. Aryabhata (Hindu) continued his work. For this ratio OPP/HYP, the word “Jya” was used

15. Brahmagupta, in 628, continued studying the same relationship and “Jya” became “Jiba” later, Jiba became Jaib, which means “fold” in Arabic

16. European Mathmeticians eventually translated “jaib” into latin: SINUS

17. Later compressed to the singular “SINE” by Edmund Gunter in 1624 Compressed again by calculator manufactorers into.. SIN

18. Given a right triangle, the 2 remaining angles must total 90 O. A = 10 O, then B = 80 O A = 30 O, then B = 60 O A B C A “compliments” B

19. O “theta” adjacent opposite hypotenuse The last ratio will be… The adjacent/hyp ratio compliments the opposite/hyp ratio (called SIN)….therefore

20. Therefore, ADJ/HYP is called “Complimentary Sinus” COSINE COS

21. The 3 Primary Trig Ratios O SINO = opp opp adj hyp COSO = adj hyp TANO = opp adj

22. Your calculator probably has hundreds of thousands of KEY triangles already loaded into the memory…..

23. O O OO Now we have 3 working ratios for every possible right sided triangle at our fingertips….

24. Solve for the following heights:

25. Finding the height of a building (H = ?) 150 m 50 O H TAN 50 = H 150 150 X TAN 50 = H X 150 150 X 1 1 178.76m = H

26. 43 O 1000 m H Tan 43 O = 1000 H X 1000 1000 X H= 1000 X Tan43 O H =932.52 m 1 1

27. 25 O 1000 m H Tan 25 O = 1000 H X 1000 1000 X H= 1000 X Tan25 O H =466.30 m 1 1

28. soh cah toa FIND A: 25 O A 17m COS25 O = A 17 X 17 17 X 1 1 A = 17 X cos25 O A = 15.4 m

29. soh cah toa FIND A: 37 O A 10 m SIN37 O = A 10 X 10 10 X 1 1 A = 10 X SIN37 O A = 6.02 m

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