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## sebastiangonzagonza one year ago HI PLEASE SOMEONE HELP ME WITH SOME QUESTIONS FOR GEOMETRY

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1. sebastiangonzagonza

@poopsiedoodle Can you please help me?

2. sebastiangonzagonza

@hero

3. mathstudent55

Here's a suggestion. Post one of the questions, and maybe someone will help you.

4. sebastiangonzagonza

can you help me with a couple please?

5. mathstudent55

Once again, post a question.

6. sebastiangonzagonza

A shooting star forms a right triangle with the Earth and the Sun, as shown below: A scientist measures the angle x and the distance y between the Earth and the Sun. Using complete sentences, explain how the scientist can use only these two measurements to calculate the distance between the Earth and the shooting star.

7. mathstudent55

Do you have the figure?

8. sebastiangonzagonza

9. sebastiangonzagonza

sorry, i was taking the screenshot of it

10. mathstudent55

No problem. I assume you are learning trigonometry of right triangles.

11. sebastiangonzagonza

yes. its one of the chapters in my online geometry course

12. mathstudent55

Ok. Here we go.

13. sebastiangonzagonza

okay. thank you so much. I really appreciate this

14. mathstudent55

In your figure, you see you have a right triangle with right angle B. The hypotenuse is AC, and that is the distance you want to find. Also, you know angle x, and you know y.

15. mathstudent55

In this triangle, the hypotenuse is AC. We already know that. We want to find this length. There are 2 legs in a right triangle. In our right triangle the legs are BC with known length y, and AB, whose length we don't know but are also not interested in.

16. sebastiangonzagonza

thats correct. we only need BC

17. mathstudent55

In a right triangle, to distinguish between the two legs, we call them the adjacent leg and the opposite leg. These terms for the legs are with reference to an acute angle of the right triangle. The angle we are using is angle C since we know its measure is x.

18. sebastiangonzagonza

so would be using SOH CAH TOA?

19. mathstudent55

Exactly. For angle C, is BC adjacent or opposite?

20. sebastiangonzagonza

I would say adjacent

21. mathstudent55

Correct. We are only interested in three things in this triangle. Known angle C (measuring x) Known adjacent leg BC (measuring y) Unknown hypotenuse AC

22. mathstudent55

Using SOH CAH TOA, which one of the trig functions sine, cosine, tangent, relates the adjacent leg to the hypotenuse?

23. sebastiangonzagonza

Cosine??

24. mathstudent55

Correct. CAH : cos A = adj/hyp

25. mathstudent55

We can write an equation using the cosine equation and our known and unknown info. $$\cos \theta = \dfrac{adj}{hyp}$$ $$\cos C = \dfrac{BC}{AC}$$ $$\cos x = \dfrac{y}{AC}$$ Since x is known, cos x is also known. y is known, so the only unknown is AC, which can be solved for. $$AC = \dfrac{y}{\cos x}$$

26. sebastiangonzagonza

so how would i put this in a essay form? this is what i have so far: To measure the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, they would be using CAH or cosine. cos A = adjacent/hypotenuse.

27. mathstudent55

Very good so far.

28. mathstudent55

Wait. I read the problem wrong.

29. mathstudent55

I'm really sorry, but let's look at it again.

30. sebastiangonzagonza

So should i write about the equations? like these: cosθ=adj/hyp cosC=BC/AC cosx=y/AC

31. sebastiangonzagonza

so we need to completely restart??

32. mathstudent55

The distance the problem is asking for is the distance between the shooting start and the Earth. We need to find AB, not AC.

33. mathstudent55

It's still a SOH CAH TOA problem only a different one.

34. sebastiangonzagonza

would they still be using cosine or would they go to tnagent?

35. sebastiangonzagonza

tagent*

36. mathstudent55

Wow, I see that you're smart. We will be using the tangent. We have x, the angle measure, and we have y, the adjacent side. The unknown is AB. AB is the opposite leg. The function that relates the opposite and adjacent legs is the tangent. TOA: tan A = opp/adj

37. mathstudent55

$$\tan x = \dfrac{opp}{adj}$$ $$\tan x = \dfrac{AB}{y}$$

38. sebastiangonzagonza

lol i try. but when i do it alone, i usually dont do too well.. so im just trying to finish out the course right now. at 12am. on my bedroom floor

39. mathstudent55

Now we solve for AB, the distance between the shooting star and the Earth.

40. mathstudent55

$$AB = y \tan x$$

41. mathstudent55

Since I started you off incorrectly, I did the following to make it up to you. I fixed your beginning of the answer. To calculate the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, he would be using TOA or tangent. tan A = opposite/adjacent.

42. sebastiangonzagonza

thank you

43. mathstudent55

Now you need to finish it.

44. sebastiangonzagonza

so here is what i have now, after correcting it: To measure the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, they would be using TOA or tangent. The equation would be opposite/adjacent so y/AB Now, you would solve for AB AB=y tan x

45. mathstudent55

You have to be careful with which side is opposite and which side is adjacent.

46. sebastiangonzagonza

so would the adjacent be C?

47. mathstudent55

To calculate the distance between the the Earth and the shooting star, the scientist would be using SOH CAH TOA. For this particular question, they would be using TOA or tangent. The equation would be opposite/adjacent so tan x = AB/y Now, you would solve for AB AB=y tan x

48. mathstudent55

Also, change "measure" to "calculate" The scientist is calculating the distance based on some measurements. He is not measuring the distance.

49. sebastiangonzagonza

alright. i changed it

50. mathstudent55

|dw:1439784419738:dw|

51. mathstudent55

Great, you got it!

52. sebastiangonzagonza

so is that all? is that everything i was supposed to write?

53. sebastiangonzagonza

@mathstudent55

54. mathstudent55

Yes. It looks good to me. You've explained how to calculate the distance from the shooting start to Earth.

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