Question

In the given figure, triangle $ABC$ is right angled at $B$. $D$ is the foot of the perpendicular from $B$ to $AC$. Given that $BC=3cm$ and $AB=4cm$. find:
i) $\tan \angle DBC$
ii) $\sin \angle DBA$

Answer 1

Let $CD=y$

then $AD=(5-y)$

and $BD=x$

In $△BCD$

${CD}^2+{BD}^2={BC}^2$ (pythagoras theorem)

$\Rightarrow y^2+x^2=9-\left(1\right)$

In $△ABD$

${AD}^2+{BD}^2=16\Rightarrow {(5-y)}^2+x^2=16-\left(2\right)$

Subtracting $\left(2\right)$ from $\left(1\right)$ we get

${(5-y)}^2-y^2=7\Rightarrow 25+y^2-10y-y^2=7\Rightarrow 10y=18\Rightarrow y=1.8\therefore CD=1.8,AD=3.2,BD=2.4$

Now,

$\left(i\right)\tan \angle DBC=\frac{CD}{BD}=\frac{1.8}{2.4}=\frac{3}{4}\left(ii\right)\sin \angle DBA=\frac{AD}{AB}=\frac{3.2}{4}=\frac{4}{5}$