A) 27.32 m
B) 25.44 m
C) 19.96 m
D) 21.92 m
Correct Answer: A
Solution :
Let AB be the building and CD be the tower. Draw \[BE\bot CD.\] Then, CE = AB = 10 m, \[\angle EBD=60{}^\circ \]and \[\angle ACB=\angle CBE=45{}^\circ \] \[\frac{AC}{AB}=\cot 45{}^\circ =1\Rightarrow \frac{AC}{10}=1\] \[\Rightarrow \] AC = 10 m From \[\Delta EBD,\]we have \[\frac{DE}{BE}=\tan 60{}^\circ =\sqrt{3}\Rightarrow \frac{DE}{AC}=\sqrt{3}\] \[\Rightarrow \] \[\frac{DE}{10}=1.732\Rightarrow DE=17.32\,\,m\] Height of the tower \[CD=CE+DE=\text{(10+17}\text{.32)}\] = 27.32 mYou need to login to perform this action.
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