Question
In figures, find the length CF.
✓
Answer
$BD = AF$
$BD = 10 cm$
In $\triangle BCD,$ we have
$\tan 30^\circ = \frac{C D}{B D}$
$\frac{1}{\sqrt{3}}=\frac{C D}{10}$
$C D=\frac{10 \times \sqrt{3}}{3} cm$
$C F=C D+D F=\frac{10 \times \sqrt{3}}{3}+2 cm$
$=\frac{10 \times \sqrt{3}+6}{3} cm $
$BD = 10 cm$
In $\triangle BCD,$ we have
$\tan 30^\circ = \frac{C D}{B D}$
$\frac{1}{\sqrt{3}}=\frac{C D}{10}$
$C D=\frac{10 \times \sqrt{3}}{3} cm$
$C F=C D+D F=\frac{10 \times \sqrt{3}}{3}+2 cm$
$=\frac{10 \times \sqrt{3}+6}{3} cm $
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