Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

Two sides of a $\Delta$le are given by the roots of the equation $x^2-2\sqrt 3x+2=0$,The angle between the sides is $\large\frac{\pi}{3}$.The perimeter of the triangle is

$\begin{array}{1 1}(a)\;6+\sqrt 3&(b)\;2\sqrt 3+\sqrt 6\\(c)\;2\sqrt 3+\sqrt{10}&(d)\;None\;of\;these\end{array}$

Can you answer this question?

1 Answer

0 votes
$x^2-2\sqrt 3x+2=0$
Angle =$\large\frac{\pi}{3}$
We know that
$\cos c=\large\frac{a^2+b^2-c^2}{2ab}$
$\cos 60^{\large\circ}=\large\frac{a^2+b^2-c^2}{2ab}$
From the given equation we have
$a+b=2\sqrt 3$
$\Rightarrow 2ab.\large\frac{1}{2}$$=a^2+b^2-c^2$
$(2\sqrt 3)^2-2\times 2-c^2=2$
$4\times 3-4-c^2=2$
$c=\sqrt 6$
Perimeter =$a+b+c$
$\qquad\;\;\;\;\;=2\sqrt 3+\sqrt 6$
Hence (b) is the correct answer.
answered Nov 18, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App