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The sides of a $\Delta$le are respectively 7cm,$4\sqrt 3$cm and $\sqrt{13}$cm then the smallest angle of the triangle is

$\begin{array}{1 1}(a)\;\large\frac{\pi}{6}&(b)\;\large\frac{\pi}{3}\\(c)\;\large\frac{\pi}{4}&(d)\;\large\frac{\pi}{5}\end{array}$

1 Answer

Let $a=7cm$
$b=4\sqrt 3$
$\Rightarrow$ Smallest side is C
Smallest angle will be C
$\cos C=\large\frac{a^2+b^2-c^2}{2ab}$
$\qquad=\large\frac{(7)^2+(4\sqrt 3)^2-(\sqrt{13})^2}{2\times 7\times 4\sqrt 3}$
$\qquad=\large\frac{49+48-13}{56\sqrt 3}$
$\qquad=\large\frac{97-13}{56\sqrt 3}$
$\qquad=\large\frac{\sqrt 3}{2}$
Hence (a) is the correct option.
answered Nov 15, 2013 by sreemathi.v

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