Question

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

then $\cos C=......$

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

Answer 1

Answer: (A)

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

$AB:BC=1:2$

So, $\frac{AB}{BC}=\frac{1}{2}=k$ ...$(k>0)$

So, $\frac{AB}{1}=\frac{BC}{2}=k$

So, $AB=k,BC=2k$

By Pythagoras theorem,

$A{C}^2=B{C}^2-A{B}^2=(2k{)}^2-k^2$
$=4k^2-k^2=3k^2$

So $AC=\sqrt{3}k$.

By definition of trigonometric ratios,
$\cos C=\frac{AC}{BC}=\frac{\sqrt{3}k}{2k}=\frac{\sqrt{3}}{2}$

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