Question
In a rectangle $\text{ABCD}, AB = 20\ cm, \angle BAC = 60^\circ ,$ calculate side $BC$ and diagonals $AC$ and $BD.$
✓
Answer
ln $\triangle A B C$
$\tan 60^{\circ}=\frac{ BC }{ AB }$
$ \Rightarrow BC =\tan 60^{\circ} \times AB$
$ \Rightarrow BC =\sqrt{3} \times 20$
$ \Rightarrow BC =20 \sqrt{3} \ cm$
$ \cos 60^{\circ}=\frac{ AB }{ AC }$
$ \Rightarrow AC =\frac{ AB }{\cos 60^{\circ}}$
$ \Rightarrow AC =\frac{20}{1}$
$ \Rightarrow AC$
$ =20 \times 2$
$ =40 \ cm$
Since diagonals of a rectangle are equal, therefore $B D=A C=40 \ cm$.
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