Question

In the $△ABC$, $AD\perp BC$ and $\angle BAD=\theta$. If $AC=20cm$, $CD=16cm$ and $BD=5cm$ then find $\sin \theta +\cos \theta$.

Answer 1

'$\theta$' is in triangle $ABD$. We need to know $AB$ and $AD$. Let us find them.
In the $△ADC$, $\angle ADC={90}^o$
$\therefore {AD}^2+{DC}^2={AC}^2$
${AD}^2+{16}^2={20}^2$
${AD}^2={20}^2-{16}^2=400-256=144$
$\therefore AD=\sqrt{144}=12cm$
Now, in $△ADB$, $\angle ADB={90}^o$
${AB}^2={AD}^2+{BD}^2={12}^2+5^2=144+25=169$
$\therefore AB=\sqrt{169}=13cm$
$\therefore \sin \theta =\frac{BD}{AB}=\frac{5}{13}$ and $\cos \theta =\frac{AD}{AB}=\frac{12}{13}$
$\therefore \sin \theta +\cos \theta =\frac{5}{13}+\frac{12}{13}=\frac{17}{13}$