Question

In $△ABC$, find the value of $\tan \angle ABC$ if $AC\perp BC$ and $\cos \angle ABC=\frac{12}{13}$.

• A. $\frac{5}{13}$
• B. $\frac{5}{12}$
• C. $\frac{12}{13}$
• D. $\frac{12}{5}$

Answer 1

The correct option is B $\frac{5}{12}$

Given: $AC\perp BC$ and $\cos \angle ABC=\frac{12}{13}$

We know that,

$\cos \angle ABC=\frac{Base}{Hypotenuse}=\frac{BC}{AB}=\frac{12}{13}$

Then, $BC=12$ and $AB=13$

In $△ABC$, by Pythagoras theorem,

$(AC{)}^2=(AB{)}^2-(BC{)}^2$

$\therefore AC=(13{)}^2-(12{)}^2=169-144=25$

$\therefore AC=5$

We know that,

$\tan \angle ABC=\frac{Perpendicular}{Base}=\frac{AC}{BC}=\frac{5}{12}$