In a $Δ A B C$ , $B C = 1$ and $A C = 2$ . Find the maximum possible value of angle A.

\frac{π}{6}

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\frac{π}{3}

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\frac{π}{2}

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\frac{π}{4}

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Solution

The correct option is A $\frac{π}{6}$
Using the cosine rule , we have
$cos θ = \frac{x^{2} + 4 - 1}{4 x}$ $= \frac{x^{2} + 3}{4 x}$
$= \frac{1}{4} [x + \frac{3}{x}]$
$= \frac{1}{4} [{(\sqrt{x} - \sqrt{\frac{3}{x}})}^{2} + 2 \sqrt{3}]$
Hence, $cos θ$ is minimum if $x = \sqrt{3}$
Therefore, the minimum value of $cos θ = 2 \times \frac{\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$
and hence maximum value of $θ = \frac{π}{6}$

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