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1-1 Using Trigonometry to Find Lengths

1-1 Using Trigonometry to Find Lengths. You have been hired to refurbish the Weslyville Tower… (copy the diagram, 10 lines high, the width of your page.). In order to bring enough gear, you need to know the height of the tower……. How would you determine the tower’s height?.

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1-1 Using Trigonometry to Find Lengths

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  1. 1-1Using Trigonometry to Find Lengths

  2. You have been hired to refurbish the Weslyville Tower…(copy the diagram, 10 lines high, the width of your page.) In order to bring enough gear, you need to know the height of the tower…… How would you determine the tower’s height?

  3. When it is too difficult to obtain the measurements directly, we can operate on a model instead. • A model is a larger or smaller version of the original object.

  4. A model must have similar proportions as the initial object to be useful. • Trigonometry uses TRIANGLES for models. We construct a similar triangle to represent the situation being examined.

  5. Imagine the sun casting a shadow on the ground. Turn this situation into a right angled triangle

  6. The length of the shadow can be measured directly The primary angle can also be measured directly X The Height? Sooo… 40O 200 m

  7. Make a model!! Draw a right angled triangle with a base of 20 cm and a primary angle of 40O, then just measure the height!

  8. X cm = 20 000 cm We can generate an equation using equivalent fractions to determine the actual height! General Model Real X cm Height 17 cm = = Base 20 000 cm 20 cm 0.85 20 000 (0.85) = X 170 m = X

  9. In the interest of efficiency.. • Drawing triangles every time is too time consuming. • Someone has already done it for us, taken all the measurements, and loaded them into your calculator • Examine the following diagram

  10. O O O O As the angle changes, so shall all the sides of the triangle. Recall the Trig names for different sides of a triangle…

  11. Geometry hypotenuse height O base Trigonometry hypotenuse opposite “theta” adjacent

  12. Trig was first studied by Hipparchus (Greek), in 140 BC. Aryabhata (Hindu) began to study specific ratios. For the ratio OPP/HYP, the word “Jya” was used

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

  14. European Mathmeticians translated “jaib” into latin: SINUS (later compressed to SIN by Edmund gunter in 1624)

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

  16. The ratio ADJ/HYP compliments the ratio OPP/HYP in the similar mathematical way. Therefore, ADJ/HYP is called “Complimentary Sinus” COSINE

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

  18. soh cah toa 1 A X 17 FIND A: 17 X COS25O = 17 1 A = 17 X cos25O 17m A = 15.4 m 25O A

  19. soh cah toa 1 A X 12 FIND A: 12 X SIN32O = 12 1 A = 12 X SIN32O 12 m A = 6.4 m A 32O

  20. soh cah toa 1 A X 10 FIND A: 10 X TAN63O = 10 1 A = 10 X TAN63O 63O A = 19.6 m 10 m A

  21. X Tan 40O = 200 200 (Tan40O) = X 168 m = X X 40O 200 m

  22. Remember: Equivalent fractions can be inverted 2 5 = 4 10 4 10 = 2 5

  23. Page 8 [1,2] a,c 3-7

  24. TAN 50 = H 150 1 150 X X 150 Find the height of the building 1 (150) TAN 50 = H HYP OPP H 150 m ADJ 50O

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