Value of an expression(in a triangle)?

**fardeen_gen** · August 1st 2008, 05:04 AM

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?

**Soroban** · August 1st 2008, 08:26 AM

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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