Question
In $\triangle ABC,$ right-angled at $B, AB = 5\ cm$ and $\angle ACB = 30^\circ .$ Determine the lengths os sides $BC$ and $AC.$
✓
Answer
Given $AB = 5\ cm $
$\angle ACB = 30^\circ$
According to diagram,
$\tan C = \frac{side \ opposite \ to \ angle \ C}{side \ adjacent \ to \ angle \ C}$
$\tan 30^\circ= \frac{AB}{BC}$
$\frac{1}{\sqrt{3}}$ = $\frac{5}{BC}$
$BC = 5\sqrt{3} cm$
$\sin C = \frac{side \ of \ angle \ C}{hypotenuse}$
$\sin 30^\circ= \frac{AB}{AC}$
$\frac{1}{2}$ = $\frac{5}{AC}$
$AC = 10\ cm.$
$\angle ACB = 30^\circ$
According to diagram,
$\tan C = \frac{side \ opposite \ to \ angle \ C}{side \ adjacent \ to \ angle \ C}$
$\tan 30^\circ= \frac{AB}{BC}$
$\frac{1}{\sqrt{3}}$ = $\frac{5}{BC}$
$BC = 5\sqrt{3} cm$
$\sin C = \frac{side \ of \ angle \ C}{hypotenuse}$
$\sin 30^\circ= \frac{AB}{AC}$
$\frac{1}{2}$ = $\frac{5}{AC}$
$AC = 10\ cm.$
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