A) 2H
B) 3H
C) H
D) None of these
Correct Answer: A
Solution :
[a] In \[\Delta \,ABD,\frac{3D}{3H}=\tan 45{}^\circ \text{ or }BD=3H\] And in \[\Delta A'BC,\frac{BC}{2H}=\tan \theta =\frac{1}{2}\text{ or }BC=H\] \[\text{Now, }y=BD-BC=3H-H=2H.\]You need to login to perform this action.
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