Question

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Answer 1

Question : As observed from the top of a 75 m high lighthouse from the sea-level, the angles of
depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships ?

Method of Solution : ↪

↪ Let AB be the lighthouse and the two ships be at point C and D respectively which are shown in the Given Figure!

↪ In ∆ABC, (Right Angled at B)

↪ Using Trigonometry Ratio!

↪ Tan 45° = AB/BC

↪ 1= 75/BC

↪ BC = 75 Cm

➡ Consider on ∆ ABD ( Right angle at B)

↪ In ∆ABD, (Right angle at B)

↪ Using Trigonometry Ratio!

↪ Tan 30° = AB/BD

↪ 1/√3 = AB/(BC+CD)

↪ 1/√3 = 75 /(75+CD)

↪ 75√3=75+CD

↪ 75√3-75=CD

↪ 75(√3-1) = CD

Or 75(1.732-1) = CD

↪ 75×0.73

↪ 54. 75 m²

Therefore, the distance between the two ships is 75(1-√3) Metres

Regards Moderator

(MasterYaduvanshi)

Answer 2

◆Solution:Let AB is the light house. So AB = 75
And C and D be the position of two ships.

From triangle ABC,

tan 45° = AB/BC

=> 1 = 75/BC

=> BC = 75

Again from triangle ADC

tan30° = AB/BD

=> 1/√3 = AB/(BC + CD) 

=> 1/√3 = 75/(75 + CD)

=> 1/√3 = 75/(75 + CD)

=> 75 + CD = 75√3

=> CD = 75√3 - 75

=> CD = 75(√3 - 1)

So the difference between ships is 75(√3 - 1) m