Question

In triangle $ABC$, $AD$is perpendicular to $BC$. $\sin B=0.8,BD=9$ cm and $\tan C=1$. Find the length of $AB,AD,AC$ and $DC$.

Answer 1

$\sin B=0.8$

So $\cos B=\sqrt{1-{\sin }^{2}B}=\sqrt{1-(0.8{)}^2}=\sqrt{1-0.64}=0.6$

In $△ABD$,

$\cos B=\frac{BD}{AB}=0.6=\frac{9}{AB}$

$AB=\frac{9}{0.6}=15$

So $AB=15cm$...(1)

$\sin B=\frac{AD}{AB}$

$0.8=\frac{AD}{15}$

$AD=12cm$...(2)

In $△ADC$,

$\tan C=\frac{AD}{CD}$

$CD=12cm$ ...(3)

and $\sec C=\sqrt{1+{\tan }^{2}C}=\sqrt{1+1^2}=\sqrt{2}$

So,

$\sec C=\frac{AC}{CD}$

$\sqrt{2}=\frac{AC}{12}$

$AC=12\surd 2cm$...(4)