A) \[\frac{\pi }{4}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{2}\]
D) \[\frac{\pi }{3}\]
Correct Answer: C
Solution :
\[A{{B}^{2}}={{\left( AD \right)}^{2}}+{{\left( BD \right)}^{2}}={{\left( 30\sqrt{3} \right)}^{2}}+{{\left( 30 \right)}^{2}}\] \[=2700+900=3600\] \[\Rightarrow \,AB=60\,\,m\]. \[\sin \alpha =\frac{1}{2}\Rightarrow \alpha =30{}^\circ \,;\,\tan \beta =\sqrt{3}\] \[\Rightarrow \beta =60{}^\circ \] \[\therefore \,\,\left| \underline{ABC} \right.+\alpha +\beta =180{}^\circ \] \[\Rightarrow \,\,\left| \underline{ABC} \right.+30{}^\circ +60{}^\circ =90{}^\circ =\frac{\pi }{2}\]
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