Question

In right angled triangle ABC, AB = 20m and $\angle CAB={60}^{\circ }$. Find BC.

• A. $\frac{20}{\sqrt{3}}$ m
• B. $10\sqrt{3}$ m
• C. $\frac{10}{\sqrt{3}}$ m
• D. $20\sqrt{3}$ m

Answer 1

The correct option is D $20\sqrt{3}$ m


In the right angle triangle ABC, AB=20m
$tan{60}^{\circ }=\frac{oppositeside}{adjacentside}=\frac{CB}{AB}=\frac{x}{20m}$
$x=tan{60}^{\circ }\times 20m$
$\Rightarrow x=20\sqrt{3}$
$(\because tan{60}^{\circ }=\sqrt{3}m)$