Question

In the figure, $△ABC$ is right angled at $B$ ans $BD$ is perpendicular to $AC$ findI)$\cos \angle CBD$II)$\tan \angle ABD$

Answer 1

Let the hypotenuse be $13$ by using Pythagoras-theorem

$\cos \left(\angle CBD\right)=\cos ({90}^{\circ }-\angle BCA)$

$=\cos \left(\angle BAC\right)$

$=\frac{AB}{AC}$

$=\frac{12}{13}$

$\tan \left(\angle ABD\right)=\tan ({90}^{\circ }-\angle BAD)$

$=\cot \left(\angle BAC\right)$

$=\frac{AB}{BC}$

$=\frac{12}{5}$