A) 50m
B) 25m
C) 46.19 m
D) None of these
Correct Answer: C
Solution :
In \[\Delta ABC\], \[\tan {{30}^{o}}\frac{BC}{40}\] \[\Rightarrow \] \[BC=\frac{40}{\sqrt{3}}\] In \[\Delta DAB,\tan {{60}^{o}}=\frac{BD}{40}\] \[\Rightarrow \] \[BD=40\sqrt{3}\] \[\Rightarrow \] \[DC+BC=40\sqrt{3}\] \[\Rightarrow \] \[DC+\frac{40}{\sqrt{3}}=40\sqrt{3}\] [from Eq. (i)] \[\Rightarrow \]\[DC=40\left( \sqrt{3}-\frac{1}{\sqrt{3}} \right)=\frac{80}{\sqrt{3}}\] \[[\because \sqrt{3}=1.73]\] \[\Rightarrow \] \[DC=46.19m\]You need to login to perform this action.
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