## anonymous one year ago i will medal anyone who helps please.its geometry Using the Tangent ratio, find BC, the height off the ground where the plank touches the wall. Round to the nearest tenth of a foot. (10 points)

1. anonymous

2. mathstudent55

The problem does tell you to use the tangent ratio. Do you know what the tangent is equal to? It is the ratio of the length of which sides of a right triangle?

3. anonymous

i dont know what its equal to no

4. mathstudent55

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5. mathstudent55

Look in the figure above. The since, cosine, and tangent are defined according to triangle side length ratios. The way to remember is SOHCAHTOA SOH: sine = opp/hyp CAH: cosine = adj/hyp TOA: tangent = opp/adj

6. mathstudent55

Ok. We see that the tangent ratio is the ratio of the lengths of the opposite leg to the adjacent leg.

7. mathstudent55

Now look in the figure with your problem. For angle C, whose measure we know, which side is the adjacent leg?

8. anonymous

im still really confused

9. mathstudent55

Do you need a more basic explanation? I can do it, I just need to find out how much I need to explain.

10. anonymous

lol very basic would help so much

11. mathstudent55

Ok. I will explain. Every time I post a response, read it, and respond ok for me to continue. The goal is for you to understand. Ok?

12. anonymous

ok

13. mathstudent55

A triangle has 3 sides. In the case of right triangle, two of the sides form a right angle. A right angle is an angle with a 90 degree measure.

14. anonymous

ok

15. mathstudent55

The three sides of a right triangles have names. The two sides that form the right angle are called "legs." The side opposite the right angle is called "hypotenuse." The hypotenuse is always the longest side of a right triangle.

16. anonymous

ok

17. mathstudent55

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18. mathstudent55

The trigonometric ratios are ratios of lengths of sides of right triangles.

19. mathstudent55

The main three trig ratios are called: sine, cosine, and tangent.

20. anonymous

ok

21. mathstudent55

Ok?

22. anonymous

ok

23. anonymous

?

24. mathstudent55

Look at the right triangle below. We are interested in angle A. |dw:1434493076172:dw|

25. anonymous

ok

26. mathstudent55

The sine ratio is defined as follows: $$\sin A = \dfrac{opposite ~leg}{hypotenuse}$$

27. anonymous

kk

28. mathstudent55

The sine of angle A is the ratio of the lengths of the opposite leg to the hypotenuse.

29. anonymous

k

30. mathstudent55

In a right triangle there is only one hypotenuse. When we say the hypotenuse, we know exactly which side we are referring to. In a right triangle, there are two legs. In general, the sides of right triangle that form the right angle are called just legs. In trig we need to distinguish between the legs, so we need to call the legs by different names to know which leg we are referring to.

31. mathstudent55

If you pick one of the acute angles of a right triangle, one leg is next to it. We call that leg the "adjacent leg" since adjacent means next to. That leg is next to that angle.

32. anonymous

k

33. mathstudent55

Once again, if you pick one of the acute angles of a right triangle, the other leg is across from that angle. We call that leg the opposite leg since opposite (in this case) means across from.

34. anonymous

ok so far im with you

35. mathstudent55

The figure below shows that for angle A, the opposite leg and the adjacent leg are positioned as labeled. |dw:1434493501483:dw|

36. anonymous

ok?

37. mathstudent55

Do you have a question?

38. anonymous

so far no

39. mathstudent55

Great. Let's continue.

40. anonymous

k

41. mathstudent55

If you use a scientific calculator, you can see that there are keys for sine, cosine, and tangent.

42. anonymous

k

43. mathstudent55

The values of the sine ratio, cosine ratio, and tangent ratio for all angles are known. Any calculator will tell you what they are.

44. mathstudent55

In trigonometry of right triangles, those same values of the the sin, cos, and tangent are ratios of lengths of sides of the right triangle.

45. anonymous

kk

46. mathstudent55

47. mathstudent55

The sine ratio is defined as the ratio of the lengths of the opposite leg to the hypotenuse. $$\sin A = \dfrac{opp}{hyp}$$

48. mathstudent55

Let's look at the following triangle and find the sine of A. |dw:1434493816047:dw|

49. mathstudent55

For angle A, the opposite leg is the one that measures 3 units. The hypotenuse measures 5 units. Since the sine ratio is defined as follows: $$\sin A = \dfrac{opp}{hyp}$$ and we know the lengths of all the sides, we can find the sine of A.

50. anonymous

i think i have a guess on my answer real quick

51. mathstudent55

For our figure above, $$\sin A = \dfrac{opp}{hyp} = \dfrac{3}{5} = 0.6$$

52. anonymous

can i run an answer by you real quick?

53. anonymous

???

54. anonymous

i believe the answer is about 3.6 but i need to put down how i got this answer

55. mathstudent55

Ok. Let's do the tangent ratio very quickly, then I'll show you how you use it.

56. anonymous

ok

57. mathstudent55

$$\tan A = \dfrac{opp}{adj}$$

58. mathstudent55

In our example above, the tangent of angle A is: $$\tan A = \dfrac{opp}{adj} = \dfrac{3}{4} = 0.75$$

59. mathstudent55

Now let's look at your problem.

60. anonymous

k

61. mathstudent55

|dw:1434494249181:dw|

62. mathstudent55

We need to use the tangent ratio to find BC, the left side of the triangle.

63. mathstudent55

The definition of the tangent ratio is: $$\tan C = \dfrac{opp}{adj}$$

64. mathstudent55

We look at angle C, and we decide which leg is adjacent and which leg is opposite.

65. anonymous

ok

66. mathstudent55

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67. anonymous

ok

68. mathstudent55

Now using the definition of the tangent ratio, and the sides of the triangle, we can write the tangent of angle C. $$\tan C = \dfrac{opp}{adj} = \dfrac{3~ft}{BC}$$ We know that angle C measures 49 deg. $$\tan 49^o = \dfrac{3}{BC}$$

69. anonymous

ok

70. mathstudent55

We want the length of BC. That is what we are trying to find. The tangent of 49 deg is simply a value that any scientific calculator can give you. We can solve the equation above for BC: $$\tan 49^o = \dfrac{3}{BC}$$ $$BC \tan 49^o = 3$$ $$BC = \dfrac{3}{\tan 49^o}$$

71. mathstudent55

Now use a calculator and divide 3 by tangent of 49 degrees.

72. mathstudent55

I get 1.1504 for tan of 49 deg, so 3/1.1504 = 2.607 That means BC = 2.6 ft

73. anonymous

ok

74. anonymous

thank you so much

75. mathstudent55

You are welcome.