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Triangle math help

  1. Jul 2, 2010 #1
    I need help....

    I am taking physics for the first time and I am taking it online which probably isnt the smartest way to go...but to make matters worse I am also currently enrolled in the calculus class that is a prerequisite for the course at the same time so my math skills are not really up to par...If anyone can help me get through this semester with a better understanding of the course material it would be greatly appreciated...Thanks

    Let the three sides of a right triangle be designated a, b, and c where c is the hypoteneuse. If c = 4, and if the angle between c and b is θ = 20°, find the lengths of the other two sides.
     
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  3. Jul 2, 2010 #2

    rock.freak667

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    Re: I need help....

    Start by drawing a diagram of the triangle and put in the angle θ.

    You should know that

    sinθ = opposite/hypotenuse and cosθ=adjacent/hypotenuse.
     
  4. Jul 2, 2010 #3
    Re: I need help....

    First off, thank you for responding so quickly...

    I drew a diagram and labeled the sides, then I used sin and cos to find the sides a and b...I got sin (20)= b/4 and cos (20)= a/4...found b= 1.3681 and a= 3.75878....the computer marked my answers wrong...
     
  5. Jul 2, 2010 #4
    Re: I need help....

    Given sinθ = 0.6, calculate tanθ without using the inverse sine function, but instead by using one or more trigonometric identities. You will find two possible values.

    I found one of the values using sin^2 (theta) + cos^2 (theta) = 1

    I tried using cos (90 + theta)= sin theta to find the second one, but couldn't remember if you were able to distribute the cos...since addition is communitive or whatever that property is called...and get cos 90 + cos theta= sin theta
     
  6. Jul 2, 2010 #5

    rock.freak667

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    Re: I need help....

    I think you put the angle in the wrong place. You need to put the angle between the sides b and c.

    Which would give you sin(20)=a/4.

    Ok well you know that sinθ is positive in quadrants 1 and 2.

    So in quadrant 1, draw a triangle at the angle θ.

    sinθ = 0.6 = 3/5 = opposite/hypotenuse

    Meaning that in your triangle, wrt θ, 3 is opposite and 5 is the hypotenuse.

    If you are unsure as to what I meant by quadrants, read http://myhandbook.info/form_trigono0.html" [Broken], the section titled "Trigonometric Functions in Four Quadrants"
     
    Last edited by a moderator: May 4, 2017
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