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# Two sides of $\Delta$le are $\sqrt 2-1$ and $\sqrt 2+1$ unit and their included angle is $60^{\large\circ}$.Then the third side of the $\Delta$le is

$(a)\;5\qquad(b)\;\sqrt 8\qquad(c)\;\sqrt 5\qquad(d)\;\sqrt 6$

We know that
$a^2=b^2+c^2-2bc\cos A$
$\;\;\;\;=(\sqrt 2-1)^2+(\sqrt 2+1)^2-2(\sqrt 2-1)(\sqrt 2+1)\cos 60^{\large\circ}$
$\;\;\;\;=2+1-2\sqrt 2+2+1+2\sqrt 2-2.(\sqrt 2)^2-(1)^2]\large\frac{1}{2}$
$\;\;\;\;=6-2(2-1).\large\frac{1}{2}$
$\;\;\;\;=6-2(1).\large\frac{1}{2}$
$\;\;\;\;=5$
$a^2=5$
$a=\sqrt 5$
Hence (c) is the correct answer.