Question: The length of the shadow is increased by 10 m on ground level of vertical building when the angle of elevation is changed from $ 45{}^\circ $ to $ 30{}^\circ . $ Find the height of the building.
Options:
A) $ 5,(\sqrt{3}+1),m $
B) $ 5,(\sqrt{3}-1),m $
C) $ 5\sqrt{3},m $
D) $ \frac{5}{\sqrt{3}},m $
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Answer:
Correct Answer: A
Solution:
- Let AB be the building and height of building be h m.
Now, in $ \Delta ABC, $
$ \tan 45{}^\circ =\frac{AB}{BC}=\frac{h}{x} $
$ \Rightarrow $ $ 1=\frac{h}{x} $
$ \Rightarrow $ $ h=x $
(i)
Again in $ \Delta ABD, $ $ \tan 30{}^\circ =\frac{AB}{BD}=\frac{h}{x+10} $
$ \frac{1}{\sqrt{3}}=\frac{h}{x+10} $
$ \Rightarrow $ $ h\sqrt{3}=x+10 $
$ \Rightarrow $ $ h\sqrt{3}=h+10 $ [from Eq.(i)]
$ \Rightarrow $ $ h\sqrt{3}-h=10 $
$ \Rightarrow $ $ h,(\sqrt{3}-1)=10 $
$ \Rightarrow $ $ h=\frac{10}{(\sqrt{3}-1)} $
$ \Rightarrow $ $ h=\frac{10(\sqrt{3}+1)}{{{(\sqrt{3})}^{2}}-{{(1)}^{2}}}=\frac{10(\sqrt{3}+1)}{2} $
$ \Rightarrow $ $ h=5,(\sqrt{3}+1) $
BETA
