## anonymous 3 years ago find x

1. anonymous
2. Mertsj

Do you know the relationships of the sides in a 30-60-90 triangle?

3. anonymous

i havect done a 30 degree one

4. anonymous

i had only done 45-90-45 i understand thoese

5. Mertsj

|dw:1359235775166:dw|

6. Mertsj

Try that. It will work.

7. anonymous

dont i devide the hypotnuse by sqrt 3?

8. anonymous

@Hero

9. Mertsj

What is the side opposite the 30 degree angle in your problem?

10. anonymous

1?

11. Mertsj

|dw:1359236251126:dw|

12. Mertsj

What is the side opposite the 30 degree angle?

13. anonymous

ohh duh 9

14. Mertsj

Now, multiply the side opposite the 30 degree angle by two and that will the hypotenuse.

15. anonymous

x=9sqrt 3

16. Mertsj

What does it mean to multiply by 2?

17. Mertsj

|dw:1359236599337:dw|

18. anonymous

so x= 18 right

19. anonymous

wow tht was easy i was over thinking it XD

20. Mertsj

yes. And if you had to find the other leg, it would be 9 multiplied by sqrt 3

21. Mertsj

$9\sqrt{3}$

22. anonymous
23. anonymous

like that will be 16sqrt 3

24. Mertsj

|dw:1359237264725:dw|

25. Mertsj

What is the side opposite the 30 degree angle labelled?

26. anonymous

y

27. Mertsj

What is the hypotenuse?

28. anonymous

16

29. Mertsj

The side opposite the 30 degree angle is 1/2 the hypotenuse. So what is y?

30. anonymous

8

31. Mertsj

And the other leg is the short leg multiplied by sqrt 3 so what is the other leg?

32. anonymous

8sqrt3

33. Mertsj

Perfect!!

34. anonymous

yas thx ur really good tutor

35. Mertsj

ty

36. anonymous

ohh yeah how can u do this

37. anonymous

|dw:1359237668453:dw|

38. anonymous

symplify

39. Mertsj

Rationalize the denominator by multipling by sqrt2 over sqrt2

40. Mertsj

$\frac{10}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

41. anonymous

10sqrt 2 is left

42. Mertsj

In the numerator, yes. What about the denominator?

43. anonymous

sqrt4

44. Mertsj

Which is?

45. anonymous

and then 2 and devide and i get 5 sqrt2

46. Mertsj

Yes. that is correct.

47. Mertsj

There is one important thing I would like for you to learn.

48. anonymous

ok

49. Mertsj

The operation "taking sqrt" is the inverse of the operation "squaring".

50. anonymous

ohh like powers 4^2=16

51. Mertsj

So when you see something like: $\sqrt{2}\times \sqrt{2}$

52. Mertsj

You should think: $(\sqrt{2})^2$

53. Mertsj

And you should think: those two operations "undo" each other so the result is 2

54. anonymous

ohh i see i get it thx

55. Mertsj

yw