Correct Answer: Option B

Explanation

TH = tan 45o, TH = QH

\(\frac{TH}{5 + QH}\) = tan 30o

TH = (b + QH) tan 30o

QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\)

QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\)

QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\)

= \(\frac{5}{\sqrt{3} - 1}\)

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