Question
Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top of the lighthouse is observed from the ships are 30º and 45º respectively. If the lighthouse is 100 m high, the distance between the two ships is:
Answer: Option C
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Let AB be the lighthouse and C and D be the positions of the ships.
Then,\(AB=100m, \angle ACB=30^{0} and \angle ADB=45^{0}
\)
\(\frac{AB}{AC}=\tan30^{0}=\frac{1}{3}\Rightarrow AC=AB\times3=1003m.\)
\(\frac{AB}{AD}=\tan45^{0}=1 \Rightarrow AD=AB=100m.\)
Therefore CD=\(\left(AC+AD\right)=\left(1003+100\right)m\)
= 100(3 + 1)
= (100 x 2.73) m
= 273 m
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