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Math Help - Trigonometry: Area of Triangles

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    Trigonometry: Area of Triangles

    Based on the books answers, (i) & (ii) are correct but I'm confused about (ii), seeing as how the answer is provided in the question!! Am I meant to prove the statement through the information provided...?
    The book's answer to (iii) is provided below, but the confusion for me here is the application of tan, in defining the area of the triangle. The 1st attempt in (iii) is what I expected the answer to be but, based on the books final answer, I made a second attempt to try and account for tan. Also, again based on the books answer, I'm assuming |ad| is meant to = r but I'm unsure as to how I prove this. Can anyone help me out here?

    Many thanks.

    Q.
    A triangle is inscribed in a sector of a circle, center c, radius r, \theta<90^o. A right-angled triangle cad circumscribes the sector, as shown. If the area of a sector is \frac{1}{2}r^2\theta, find the area of (i) \triangle cab, (ii) the sector cab, (iii) \triangle cad. Hence show that sin\theta<\theta<tan\theta.

    Attempt: (i) Area of \triangle abc: \frac{1}{2}(a)(b)(sinC)
    |ac| = r = |bc|, where both lines are connected from the center to the circumference of the circle whose sector is cab
    Thus, \frac{1}{2}(a)(b)(sinC) = \frac{1}{2}r^2sin\theta

    (ii) \frac{1}{2}r^2\theta

    (iii) Area of \triangle cad: \frac{1}{2}(d)(a)(sinA) = \frac{1}{2}(r)(|cd|)(sin90^o) = \frac{1}{2}(r)(a)(1) = \frac{1}{2}ra
    or...?
    \frac{1}{2}(c)(d)(tan\theta) = \frac{1}{2}(|ab|)(r)(tan\theta)

    Ans.: (From text book): (iii)
    \frac{1}{2}r^2tan\theta
    Attached Thumbnails Attached Thumbnails Trigonometry: Area of Triangles-photo-2-.jpg  
    Last edited by GrigOrig99; August 3rd 2012 at 01:55 PM.
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  2. #2
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    Re: Trigonometry: Area of Triangles

    note that |ad| = r\tan{\theta}

    A = \frac{1}{2}bh = \frac{1}{2} r \cdot r\tan{\theta}
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    Re: Trigonometry: Area of Triangles

    Thanks.

    Ok, so if I'm to demonstrate sin\theta<\theta<tan\theta, and knowing that \theta<90^o, I can simply pick any degree from 0^0-89^0 and display the results e.g. \theta=60^0 in which case sin\60^o<\60^o<tan\60^0 = 0.86<60<1.73. Although, this leaves me with the problem of the 60 in the middle.

    Is there a better way to prove the statement?
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    Re: Trigonometry: Area of Triangles

    Quote Originally Posted by GrigOrig99 View Post
    Thanks.

    Ok, so if I'm to demonstrate sin\theta<\theta<tan\theta, and knowing that \theta<90^o, I can simply pick any degree from 0^0-89^0 and display the results e.g. \theta=60^0 in which case sin\60^o<\60^o<tan\60^0 = 0.86<60<1.73. Although, this leaves me with the problem of the 60 in the middle.
    Is there a better way to prove the statement?
    \text{Area}(\Delta CAB)\le \text{Sector-area}(CAB)\le\text{Area}(\Delta CAD)
    Thanks from GrigOrig99
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    Re: Trigonometry: Area of Triangles

    Ok, thank you.
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