Question

In $△ABC,m\angle A={90}^{\circ }$ and if $AB:BC=1:2$, then $\sin B=.....$

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $\sqrt{3}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}}{2}$

$AB:BC=1:2,\therefore \frac{AB}{AC}=\frac{1}{2}$

$\frac{AB}{1}=\frac{AC}{2}=k$ ...$(k>0)$

So, $AB=k,AC=2k$

By Pythagoras theorem,

$A{C}^2=B{C}^2-A{B}^2=(2k{)}^2-(k{)}^2=4k^2-k^2$

$A{C}^2=3k^2$, So $AC=\sqrt{3}k,AB=k,BC=2k$

$\sin B=\frac{AC}{BC}=\frac{\sqrt{3}K}{2k}=\frac{\sqrt{3}}{2}$ ....(By definition of trigonometric ratios)

So, $\frac{\sqrt{3}}{2}$ is correct.