Question
Find:$a. BC;b. AD;c. AC$
✓
Answer
$a.$ In right $\triangle A B C$,
$\tan 30^{\circ}=\frac{ AB }{ BC }$
$ \Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{ BC }$
$ \Rightarrow BC =10 \sqrt{3} \ cm .$
$b.$ In $\triangle A B C, \angle C=30^{\circ}$ and $\angle B=90^{\circ}$
$\Rightarrow \angle A=60^{\circ}$
Now, In $\triangle ABD$,
$\cos 60^{\circ}=\frac{A D}{A B}$
$ \Rightarrow \frac{1}{2}=\frac{A D}{10}$
$ \Rightarrow A D=5 \ cm$
$ c$.In $\triangle A B C$
$ A C^2$
$ =A B^2+B C^2$
$ =10^2+(10 \sqrt{30})^2$
$ =100+300$
$ =400 \ cm$
$ \Rightarrow A C=20 \ cm .$
$\tan 30^{\circ}=\frac{ AB }{ BC }$
$ \Rightarrow \frac{1}{\sqrt{3}}=\frac{10}{ BC }$
$ \Rightarrow BC =10 \sqrt{3} \ cm .$
$b.$ In $\triangle A B C, \angle C=30^{\circ}$ and $\angle B=90^{\circ}$
$\Rightarrow \angle A=60^{\circ}$
Now, In $\triangle ABD$,
$\cos 60^{\circ}=\frac{A D}{A B}$
$ \Rightarrow \frac{1}{2}=\frac{A D}{10}$
$ \Rightarrow A D=5 \ cm$
$ c$.In $\triangle A B C$
$ A C^2$
$ =A B^2+B C^2$
$ =10^2+(10 \sqrt{30})^2$
$ =100+300$
$ =400 \ cm$
$ \Rightarrow A C=20 \ cm .$
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