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Find the slope and angle of inclination

  1. Feb 16, 2005 #1
    1) Find the slope
    a) (3.14)/6 b) 135° c)60°

    2) Find the angle of inclination of a line with the given slope
    a) m=-1/2 b) m=1 c) m=-2 d) m=57

    how do u do these questions?
  2. jcsd
  3. Feb 16, 2005 #2


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    1) {Slope} = tan{Angle of Inclination}
    2) {Angle of Inclination} = tan^(-1){Slope}

    a) tan^(-1) = arctan = {Inverse tan}
    b) In #1 above, determine the units of each "Angle of Inclination" before doing calculations and be sure calculator is set to the proper units (e.g., degrees or radians).

    Last edited: Feb 16, 2005
  4. Feb 16, 2005 #3
    Hello, gillgill, I'll do a few for you:

    The slope of a line passing through points [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] is given by [tex]m=y_2-y_1/x_2-x_1 = \tan\alpha[/tex], where [tex]\alpha[/tex] is the angle between the line and the x-axis.

    So, if given an angle [tex]\alpha=\pi/6=180^o/6=30^o[/tex], you find the slope by evaluating [tex]m=\tan30^o[/tex], which I believe is [tex]m=\sqrt{3}/3[/tex].

    Now, if given a slope [tex]m=-1/2[/tex], you know that [tex]m=-1/2=\tan\alpha[/tex]. If you know the value of [tex]\tan\alpha[/tex] (in this case [tex]-1/2[/tex]), then you can find [tex]\alpha[/tex], by evaluating [tex]\arctan m=\arctan(-1/2)=\alpha[/tex], which gives you an angle of inclination [tex]\alpha=-26.57^o[/tex].

    Hope this helps! Now try to do the other ones by yourself.:smile:

    - Kamataat
  5. Feb 16, 2005 #4
    but for ex. 1a) the answer is 1/√3...how do u find that out?
  6. Feb 16, 2005 #5
    o..okok..i see now....thx guys
  7. Feb 16, 2005 #6

    - Kamataat
  8. Feb 16, 2005 #7
    how do u go from tan30 to 1/√3?
  9. Feb 16, 2005 #8


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    Things you should know:

    sin 30 = cos60 = 1/2
    cos 30 = sin 60 = √3/2
    tan 30 = sin30/cos30 = 1/2 / √3/2 = 1/√3

    You should also know that sin45 and cos45 are both √2/2. And therefore tan45 is 1.

    It is also helpful to remember approximate values for √3/2 and √2/2. They are .866 and .707 respectively. I can remember in high school my knowledge of those approximations helped me answer many questions I wasn't exactly sure of, quite quickly.
  10. Feb 16, 2005 #9
  11. Feb 16, 2005 #10


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    If u're to remember someting,try to remember [itex] \sqrt{2}\approx 1.414 [/itex] and [itex]\sqrt{3}\approx 1.732 [/itex] or maybe with 3 sign.figures,only...

    That way,u can do whatever operations with them.

  12. Feb 17, 2005 #11
    i have a different question this time...
    given y=√x-4
    reflected in the y-axis
    translated 5 units left
    compressed horizontally by 1/3

    i keep getting y=√-3(x+1)...which is wrong...can anybody tell me what my mistake is?
  13. Feb 17, 2005 #12
    How can I type latex faster?>
  14. Feb 17, 2005 #13


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    How in the world did you get that negative??

    In general, a horizontal (left-right) change is a change in x.

    Translating 5 units left shifts means that "x= -5" acts like "x= 0": that is the formula must involve "x+ 5". (Shifting 0 to a is the same as replacing x by x-a: in this case a= -5.)

    "Compressing horizontally by 1/3" means "x= 1/3" acts like "x=1". That will be true if we multiply x by 3: replace x by 3x.

    Putting those together, we can "translate 5 units left" and then "compress horizontally by 1/3" by replacing x by 3(x+ 5).

    Since the original function was y= √(x-4) (I am assuming the entire x-4 was inside the root), the new function will be √(3(x+5)-4)=
    √(15x+ 11).
  15. Feb 17, 2005 #14
    does reflect in the y-axis gives x an negative?
    by the way...the answer is y=√-3(x+3)
  16. Feb 17, 2005 #15
    This is what I did:

    Original function: [tex]y=\sqrt{x-4}[/tex]

    Reflect on the y-axis: [tex]y=\sqrt{-x-4}[/tex]

    Translate horizontally 5 units to the left: [tex]y=\sqrt{-(x+5)-4}=\sqrt{-x-5-4}=\sqrt{-x-9}[/tex]

    Compress horizontally by 1/3: [tex]y=\sqrt{-3x-9}=\sqrt{-3(x+3)}[/tex]

    - Kamataat
  17. Feb 17, 2005 #16
    o..i get it now...thx
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