Question
Two ships are on either side of a 100 m lighthouse. The angles of elevation from the two ships to the top of the lighthouse are 30° and 45°. Find the distance between the ships.Â
Answer: Option C
:
C
Let, BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m, ∠BAD = 30∘ , ∠BCD = 45∘
tan30∘=BDBA⇒1√3=100BA⇒BA=100√3tan45∘=BDBC⇒1=100BC⇒BC=100
Distance between the two ships
=AC=BA+BC=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273m
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:
C
Let, BD be the lighthouse and A and C be the positions of the ships.
Then, BD = 100 m, ∠BAD = 30∘ , ∠BCD = 45∘
tan30∘=BDBA⇒1√3=100BA⇒BA=100√3tan45∘=BDBC⇒1=100BC⇒BC=100
Distance between the two ships
=AC=BA+BC=100√3+100=100(√3+1)=100(1.73+1)=100×2.73=273m
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