$△ A B C,$ if $∠ C = 3 ∠ A, B C = 27,$ and AB=48. Then the value of $A C$ is ___________

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Solution

Given $a = 27, c = 48$
Using sine rule, $\frac{a}{sin A} = \frac{c}{sin 3 A}$ (since $C = 3 A$ )
$\Rightarrow \frac{27}{sin A} = \frac{48}{3 sin A - 4 {sin}^{3} A}$
Solving this we get, $sin A = \frac{\sqrt{11}}{6}$

$∴ sin C = \frac{8 \sqrt{11}}{27}$
Hence $b = a sin C + c sin A = 45$

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