In a triangle ABC, $B = 30^{o}, c = 150, b = 50 \sqrt{3}$
$a_{1}, a_{2}$ are the two values of the third side $a$ . $x$ and $y$ are respectively the arithmetic and geometric means of $a_{1}$ and $a_{2}$ then ${(\frac{x}{\sqrt{3}})}^{2} - {(\frac{y}{\sqrt{6}})}^{2}$ is equal to?

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Solution

$b^{2} = a^{2} + c^{2} - 2 a c c o s B$
$\Rightarrow (50 \sqrt{3})^{2} = a^{2} + (150)^{2} - 2 a . 150 (\sqrt{3} / 2)$
$\Rightarrow a^{2} - 150 \sqrt{3} a + 15000 = 0$
$\Rightarrow a = 50 \sqrt{3}$ or $100 \sqrt{3}$
Let $a_{1} = 50 \sqrt{3}, a_{2} = 100 \sqrt{3}$
Now, $x = \frac{a_{1} + a_{2}}{2}$
$\Rightarrow x = 75 \sqrt{3}$
Also, $y^{2} = a_{1} a_{2}$
$\Rightarrow y = 50 \sqrt{6}$
So, ${(\frac{x}{\sqrt{3}})}^{2} + {(\frac{y}{\sqrt{6}})}^{2}$
$= (75)^{2} - (50)^{2} = 3125$

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