Using the given figure, BC = ___ cm.

$12 \sqrt{3}$

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$4 \sqrt{3}$

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Solution

The correct option is A
$12 \sqrt{3}$

$t a n 30^{\circ} = \frac{A B}{B C}$

$\frac{1}{\sqrt{3}} = \frac{12}{B C}$

$B C = 12 \sqrt{3} c m$

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Similar questions

In the given figure, AB = 5 cm, BC = 4 cm, AC = 3 cm. The diameter of the circle is

B C = 15

sin B = \frac{4}{5}

A C

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