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From the figure, calculate TH in centimeters


Question

From the figure, calculate TH in centimeters

Options

A) \(\frac{5}{\sqrt{3} + 1}\)

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

C) \(\frac{5}{\sqrt{3}}\)

D) \(\frac{\sqrt{3}}{5}\)

The correct answer is B.

Explanation:

TH = tan 45\(^{\circ}\), TH = QH

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

TH = (b + QH) tan 30\(^{\circ}\)

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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Dicussion (1)

  • TH = tan 45\(^{\circ}\), TH = QH

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

    TH = (b + QH) tan 30\(^{\circ}\)

    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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