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Math Help - trignometry

  1. #1
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    trignometry

    the hypotenuse of a right-angled triange has the length of 10cm. find the expression for the area of the triangle in terms of theta. for what value of theta is the area a maximum and what is this maximum area?
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  2. #2
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    Hello,

    Though you didn't define theta, I assume it is the value of one of the angles which is not the right angle.

    The length of the sides about the right angle are 10sin(theta)cm and 10cos(theta)cm. Now you can compute the area.
    For its maximum, use the AM-GM inequality \sqrt{\sin^2\theta\cos^2\theta}\leq \frac{\sin^2\theta+\cos^2\theta}{2}.

    Bye.
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  3. #3
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    Hello, anitha!

    The hypotenuse of a right-angled triange has the length of 10cm.
    Find an expression for the area of the triangle in terms of \theta
    For what value of \theta is the area a maximum and what is this maximum area?

    A right triangle can be inscribed in a semicircle.
    Code:
                * * *    C
            *           *
          *         *   :** 
         *      *      h: **
            * θ         :  *
      A * - - - - - - - + - * B
                 10

    The base of the triangle is 10.

    AC \:=\:10\cos\theta \quad\Rightarrow\quad h \:=\:(10\cos\theta)(\sin\theta)

    Hence, the area is: . A \;=\;\frac{1}{2}(10)(10\sin\theta\cos\theta) \;=\;50\sin\theta\cos\theta

    Therefore: . \boxed{A \;=\;25\sin2\theta}



    The area is a maximum when h is a maximum.
    . . This occurs when  h = 5 (radius) . . . \theta = 45^o

    Therefore, the maximum area is: . A \;=\;\frac{1}{2}(10)(5) \;=\;\boxed{25\text{ cm}^2}

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  4. #4
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    Quote Originally Posted by anitha View Post
    the hypotenuse of a right-angled triange has the length of 10cm. find the expression for the area of the triangle in terms of theta. for what value of theta is the area a maximum and what is this maximum area?
    Then, the leg that is opposite theta is 10sin(t)......t is theta
    And, the leg adjacent to theta is 10cos(t)
    So, area of right triangle, A =(1/2)[10sin(t)[10cos(t)]
    A = 50sin(t)cos(t) ----------answer.
    Or if 2*t is allowed,
    A = 25sin(2t) ---------------- could be an answer.

    ----------------------
    For the maximum area.

    a) Are you allowed to use Calculus?

    0 = 50cos(2t)
    0 = cos(2t)
    2t = arccos(0) = 90deg or 270 deg.....Or, pi/2 or 3pi/2 radians
    So,
    t = 45deg or 135 deg
    t cannot be 135 deg because any triangle can only have 180 degees for its 3 interior angles.
    Hence t = 45deg or pi/4 radians for maximun area ------answer.

    A = 50sin(t)cos(t)
    So,
    max A = 50sin(pi/4)cos(pi/4)
    max A = 50[1/sqrt(2)][1/sqrt(2)] = 25 sq.cm ------------answer.


    --------
    b) if you are not allowed to use Calculus, then draw the graph of A = 25sin(2t).
    Then find when A is maximum.

    A = 25sin(2t) -------(i)
    amplitude is 25
    period is 2pi/2 = pi radians
    there are no horizontal or vertcal shifts.
    As a sine curve, its maximum occurs at 1/4 of the period.
    That is at t = (1/4)pi radians for max A ------answer.
    So,
    max A = 25sin[2*(1/4)pi]
    max A = 25sin(p/2)
    max A = 25(1) = 25 sq.cm -----------answer.
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