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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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For a $\Delta ABC$ it is given that : $\cos A+\cos B+\cos C=\large\frac{3}{2}$.What type of triangle is $ABC$?

$\begin{array}{1 1}(a)\;Isosceles \;triangle&(b)\;Right\;angled\;triangle\\(c)\;equilateral\;triangle&(d)\;None\;of\;these\end{array}$

Can you answer this question?
 
 

1 Answer

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Step 1:
Given: $\cos A+\cos B+\cos C=\large\frac{3}{2}$
$\Rightarrow \large\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}=\frac{3}{2}$
$\Rightarrow a(b^2+c^2-a^2)+b(c^2+a^2-b^2)+c(a^2+b^2-c^2)=3abc$
$\Rightarrow a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)=a^3+b^3+c^3+3abc$
$\Rightarrow a(b^2+c^2-2bc)+b(c^2+a^2-2ac)+c(a^2+b^2-2ab)=a^3+b^3+c^3-3abc$
$\Rightarrow a(b-c)^2+b(c-a)^2+c(a-b)^2-\large\frac{1}{2}$$(a+b+c)[(b-c)^2+(c-a)^2+(a-b)^2]=0$
$\Rightarrow (b-c)^2(b+c-a)+(c-a)^2(c+a-b)+(a-b)^2+(a+b-c)=0$
Since in a triangle, the sum of any two sides is always > the third side,
Step 2:
All terms in LHS are non negative.
$\Rightarrow\:$Each term =0
$\Rightarrow b-c=c-a=a-b=0$
$a=b=c$
$\Delta$ABC is a equilateral triangle.
Hence (c) is the correct answer.
answered Nov 19, 2013 by sreemathi.v
edited Mar 25, 2014 by rvidyagovindarajan_1
 

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