## sarver1995 3 years ago in a 45 45 90 triangle i need to find what x is

1. sarver1995

|dw:1342917395925:dw|

2. Romero

|dw:1342917641735:dw|

3. sarver1995

would it be 6 square root of 2

4. Romero

Use Pythagoras theorem $x^2+x^2=12^2$

5. Romero

yes

6. Australopithecus

you cant solve this with pythagoras theorem you have to use trig I'm pretty sure

7. Romero

$2x^2=12^2$$2x^2=144$$x^2=\frac{144}{2}$$x^2=72$$x= \sqrt{72}$$x= \sqrt{36*2}$$x=6 \sqrt{2}$

8. Romero

Yes you can you have two sides that are equal

9. Calcmathlete

Hey @sarver1995 Been a while :) 45-45-90 triangles: |dw:1342917784578:dw|

10. Australopithecus

oh yeah just noticed :)

11. Calcmathlete

$x\sqrt{2} = 12$$x = \frac{12}{\sqrt{2}}$$x = \frac{12}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$$x = \frac{12\sqrt{2}}{2}$$x = 6\sqrt{2}$There you go :)

12. Australopithecus

SOH CAH TOA Sin = opposite/hypotenuse cos = adjacent/hypotenuse tan = opposite/adjacent cos(45) = x/12 cos(45)(12) = x

13. Romero

from this I noticed that you can simply get one of the sides by multiplying the hypotenuse by $\frac{1}{\sqrt{2}}$

14. Romero

Only if the triangle is a 45 45 90

15. sarver1995

|dw:1342917788335:dw|

16. Calcmathlete

Same thing...use the ratio above... |dw:1342918025299:dw|

17. Romero

same thing like I said $\frac{1}{\sqrt{2}}*x=10$

18. Romero

solve for x

19. sarver1995

to calcmathlete yeah i am now reviewing and oh course still confused i only have a few more lessons and i am finished with 11th grade. its been really rough

20. sarver1995

|dw:1342918127585:dw| would x be 10 square root of 2

21. Calcmathlete

Yes :) $x = 10\sqrt{2}$

22. Romero

Good job mate!

23. sarver1995

|dw:1342918257888:dw| okay not sure how to do this one

24. sarver1995

it cant be 1/2 square root of 2

25. Calcmathlete

Does it tell you if it's a right triangle?

26. sarver1995

45 45 90

27. Calcmathlete

It tells you that?

28. sarver1995

yes

29. Calcmathlete

Ok. The ratio again. x, x, x√2 x = 1.2 x√2 = ?

30. sarver1995

i will try to draw it again maybe i ddin't do it right

31. Calcmathlete

By the way, if the x's get confusing because they are using x as a variable, I'll just use y.

32. sarver1995

|dw:1342918455394:dw|

33. Calcmathlete

Ok. Same situation. The legs are always congruent in a 45-45-90 triangle. The hypotenuse is always in this ratio: $Leg \times \sqrt{2}$Can you figure it out since you have the leg?

34. sarver1995

1.2 x square root of 2

35. sarver1995

not sure on this one

36. Calcmathlete

Yup. which can be rewritten as $$1.2\sqrt{2}$$

37. sarver1995

okay i have one more like these can i try it with us

38. Calcmathlete

Alright shoot away :)

39. sarver1995

|dw:1342918731847:dw|

40. Calcmathlete

$x = \frac{\sqrt{24}}{\sqrt{2}} = \sqrt{\frac{24}{2}} = \sqrt{12} = ?$

41. sarver1995

2 square root of 3

42. Calcmathlete

yup :)

43. sarver1995

okay i have a different one i just need u to help me understand the problem it will take a little while to draw

44. Calcmathlete

ok.

45. sarver1995

|dw:1342918993022:dw| Refer to the figure to complete the following item. Given: If m arc vu =80° and marc st =40°, then 1 =

46. Calcmathlete

Oh gosh...a proof?

47. sarver1995

yeah lol

48. Calcmathlete

Do you have to like fill in a blank or is it like write an entire proof?

49. sarver1995

its just asking what angle 1 would equal

50. sarver1995

51. Calcmathlete

I think that it's 50...

52. sarver1995

can you tell me how you got that because i have three more problems like this one

53. Calcmathlete

$m\angle1 = \frac 12(arcVT - arcST)$

54. sarver1995

If arc vu m = 70° and m arc st = 30°, then 2 = would it be the same formula for this one

55. Calcmathlete

I kind of have to go...sorry :/ If it maintains the same shape, it should be the same sort of formula...

56. sarver1995

thanks for your help its really appreciated and God bless you