| In the given figure, ABCD is a rectangle with \[AD=8cm\]and \[CD=12cm\]. Line segment CE is drawn, making an angle of \[60{}^\circ \]with AB, intersecting AB at E. Find the length of CE and BE respectively. |
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A) \[\frac{16}{\sqrt{3}}cm,\,\,\frac{8}{\sqrt{3}}cm\]
B) \[\frac{8}{\sqrt{3}}cm,\,\,\frac{16}{\sqrt{3}}cm\]
C) \[\frac{16}{\sqrt{3}}cm,\,\,\frac{7}{\sqrt{3}}cm\]
D) \[\frac{8}{\sqrt{3}}cm,\,\,\frac{5}{\sqrt{3}}cm\]
Correct Answer: A
Solution :
| [a] In right angled \[\Delta EBC,\] |
| \[\sin 60{}^\circ =\frac{BC}{CE}\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\frac{\sqrt{3}}{2}=\frac{AD}{CE}\]\[[BC=AD]\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\frac{\sqrt{3}}{2}=\frac{8}{CE}\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,CE=\frac{16}{\sqrt{3}}cm\] .(1) |
| Also,\[\cos 60{}^\circ =\frac{BE}{CE}\Rightarrow \,\,\,\frac{1}{2}=\frac{BE}{\left( \frac{16}{\sqrt{3}} \right)}\] [From eq.(1)] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,2BE=\frac{16}{\sqrt{3}}\,\,\,\,\Rightarrow \,\,\,BE=\frac{8}{\sqrt{3}}cm\] |
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