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Math Help - Value of an expression(in a triangle)?

  1. #1
    Super Member fardeen_gen's Avatar
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    Value of an expression(in a triangle)?

    If in a triangle ABC a:b:c = 2:4:5, then the value of (8.cosB.cosC + 64.cos C.cos A + 125.cos A.cos B)/(1 - 2.cos A.cos B.cos C) is-

    A) 20
    B) 35
    C) 40
    D) None

    I am getting none as the answer. Is it correct?
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  2. #2
    Super Member

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    Hello, fardeen_gen!

    If in a triangle ABC,\;\;a:b:c \,=\, 2:4:5,

    find the value of: . X \;=\;\frac{8\cos B\cos C + 64\cos C\cos A + 125\cos A\cos B}{1 - 2\cos A\cos B\cos C}

    . . A)\;20\qquad B)\;35 \qquad C)\;40 \qquad D) \text{ None}
    Let: . \begin{array}{ccc}a &=&2k \\ b &=& 4k \\ c &=&5k\end{array}


    Law of Cosines:. \cos A \;=\;\frac{b^2+c^2-a^2}{2bc}


    \cos A \:=\: \frac{b^2+c^2-a^2}{2bc} \:=\: \frac{16k^2 + 25k^2-4k^2}{2(4k)(5k)} \:=\: \frac{37k^2}{40k^2} \:=\: \frac{37}{40}

    \cos B \:=\:\frac{a^2+c^2-b^2}{2ac} \:=\:\frac{4k^2+25k^2-16k^2}{2(2k)(5k)} \:=\:\frac{13k^2}{20k^2} \:=\:\frac{13}{20}

    \cos C \:=\:\frac{a^2+b^2-c^2}{2ab} \:=\:\frac{4k^2+16k^2-25k^2}{2(2k)(4k)} \:=\:\frac{\text{-}5k^2}{16k^2} \:=\:\text{-}\frac{5}{16}


    Then: . X \;=\;\frac{8\left(\frac{13}{20}\right)\left(\text{-}\frac{5}{16}\right) + 64\left(\frac{37}{40}\right)\left(\text{-}\frac{5}{16}\right) + 125\left(\frac{27}{40}\right)\left(\frac{13}{20}\r  ight)} {1 - 2\left(\frac{37}{40}\right)\left(\frac{13}{20}\rig  ht)\left(\text{-}\frac{5}{16}\right)}

    . . . . . . . = \;\frac{-\frac{13}{8} - \frac{37}{2} + \frac{2405}{32}} {1 + \frac{2405}{6400}} \;=\;\frac{\frac{1761}{32}}{\frac{8805}{6400}}

    . . . . . . . = \;\frac{1761}{32}\cdot\frac{6400}{8805} \;=\;{\color{blue}40}\quad\hdots\;\;{\color{blue}\  text{ answer (C)}}


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