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^o$
According to diagram,
tan C = $\frac{side \ opposite \ to \ angle \ C}{side \ adjacent \ to \ angle \ C}$
tan $30^o$ = $\frac{AB}{BC}$
$\frac{1}{\sqrt{3}}$ = $\frac{5}{BC}$
BC = 5$\sqrt{3}$ cm
sin C = $\frac{side \ of \ angle \ C}{hypotenuse}$
sin $30^o$ = $\frac{AB}{AC}$
$\frac{1}{2}$ = $\frac{5}{AC}$
$AC = 10 cm.$
$\angle$$ACB = 30^o$
According to diagram,
tan C = $\frac{side \ opposite \ to \ angle \ C}{side \ adjacent \ to \ angle \ C}$
tan $30^o$ = $\frac{AB}{BC}$
$\frac{1}{\sqrt{3}}$ = $\frac{5}{BC}$
BC = 5$\sqrt{3}$ cm
sin C = $\frac{side \ of \ angle \ C}{hypotenuse}$
sin $30^o$ = $\frac{AB}{AC}$
$\frac{1}{2}$ = $\frac{5}{AC}$
$AC = 10 cm.$
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