A) BC : 2 CE
B) 2 CE : BC
C) 2 BC : CE
D) CE : 2 BC
Correct Answer: A
Solution :
[a] Given, \[\tan \,\angle ABC=3.6\] In \[\Delta DCE,\tan \angle DCE=\frac{18}{5}=3.6\] \[\therefore \]\[\Delta ABC\] is an isosceles triangle where \[AB=AC\] Now, drawing \[AM\bot BC\] \[DE\parallel AM,\] \[\therefore \]In \[\Delta ACM\]and \[\Delta DCE\] \[\frac{AC}{CD}=\frac{MC}{CE}=\frac{\frac{1}{2}BC}{CE}=\frac{BC}{2CE}\] |
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