In a rectangle $ A B C D, A B=20 \mathrm{~cm}, \angle B A C=60^{\circ} $, calculate side $ B C $ and diagonals $ A C $ and BD.

Given:

In a rectangle $ A B C D, A B=20 \mathrm{~cm}, \angle B A C=60^{\circ} $.

To do:

We have to calculate side $ B C $ and diagonals $ A C $ and BD.

Solution:

$ \mathrm{ABCD} $ is a rectangle in which $ \mathrm{AB}=20 \mathrm{~cm}, \angle \mathrm{BAC}=60^{\circ} $
$ \angle B=90^{\circ} $
$ \cos \angle \mathrm{BAC}=\frac{\mathrm{AB}}{\mathrm{AC}} $
$ \Rightarrow \cos 60^{\circ}=\frac{20}{\mathrm{AC}} $

$ \Rightarrow \frac{1}{2}=\frac{20}{\mathrm{AC}} $ (Since $\cos 60^{\circ}=\frac{1}{2})$)
$ \Rightarrow A C=20 \times 2=40 \mathrm{~cm} $
$ \tan \angle \mathrm{BAC}=\frac{\mathrm{BC}}{\mathrm{AB}} $

$ \Rightarrow \tan 60^{\circ}=\frac{\mathrm{BC}}{20} $
$ \Rightarrow \sqrt{3}=\frac{\mathrm{BC}}{20} $ (Since $\tan 60^{\circ}=\sqrt3)$)

$ \Rightarrow \mathrm{BC}=20 \sqrt{3} \mathrm{~cm} $

The side $ \mathrm{BC} $ is $ 20 \sqrt{3} \mathrm{~cm} $ and the diagonals $ AC $ and $ BD $ are $ 40 \mathrm{~cm} $.

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Updated on: 10-Oct-2022

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