Two ships are sailing in the sea on the two sides of a lighthouse. The angle of elevation of the top

M4maths Previous Puzzles

29 Aug 2017 Distance

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 45° and 60° respectively. If the lighthouse is 222 m high, the distance between the two ships is:

OPtion

1) 350.17 m
2) 352.17 m
3) 353.17 m
4) 354.17 m
5) 355.17 m
6) 356.17 m
7) 357.17 m
8) 358.17 m
9) 359.17 m
10)None of these

Solution

Let AB be the lighthouse where A is the bottom point of light house and C and D be the positions of the ships.
Then, AB = 222 m, Angle ACB = 45° and angle ADB = 60°.
AB/AC = tan 45° = 1
=> AC = AB = 222

AB/AD = tan 60° = 3^(1/2)
=> AD = AB/3^(1/2) = 74*3^(1/2)

CD = (AC + AD) = 222+ 74*3^(1/2)
= 350.17
Option 1)

Correct Option: 1 Best Solution (15)

AYAN PAUL   7 years AGO

The ships are sailing on both sides of the lighthouse
So the distance between the Lighthouse and ship which makes the angle 45 degree will be = 222 m as tan 45 = 1
Similarly the distance between the lighthouse and the other ship will be 222/1.732 = 128.17 m
So the distance between the two ships will be = 222+128.17 = 350.17 m

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G.S.Kantharao   7 years AGO

Angles of elevations =45°,60° respectively.
Let the distances from the light house are = x,y
Tan 45° = 222/x, Tan 60° = 222/y
1 = 222/x, √3 = 222/y
The distance between the ship's = x + y = 222 + 128.1717
=350.1717..
Hence the right option is 1.👈.

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SAYAN CHATTERJEE   7 years AGO

Distance between two ships is 222*(1/tan45+1/tan60)
= 350.17.
Option 1.

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Praveenkumar P Kalikeri   7 years AGO

Distance
= tan(45) + (1/sqrt(3))*(222)
= 222.00+128.171=350.171

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mukesh   7 years AGO

to find the distance between two ships.
by using the trignometric method we can solve this.
tan45= 222/ab,ab=222m.
tan60=222/bc,bc=128.17m
therefore.the distance between two ships is,ab+bc=222+128.17=350.17m

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Varun   7 years AGO

It is 222/tan45 + 222/tan(60)

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Mukund Manohar   7 years AGO

Distance between ships=222 + 222/sq.root3
= 350.17m

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Suraj Bhushan Pandey   7 years AGO

As the height and angle of elevation are given from ships so simply the tan trigonometry formula

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KOWSALYA   7 years AGO

tan45=222/x=>x=222
tan60=222/y=>y=128.17
x+y=350.17

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shraboni   7 years AGO

the angle of elevation is 45°and 60°
let BC is the lighthouse =222m, AC is the distance between the 1st ship and lighthouse, also CD is the distance from the 2nd ship.we have to find AD.

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Anupam Tripathi   7 years AGO

height of lighthouse=222
tan45=222/base1
base1=222
now, again
tan60=222/base2
base2=222/1.732=128.175
now, total base=base1+base2=350.17m

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yamini velaga   7 years AGO

let the distances of the ships from light house be x & y respectively
height of light house = 222m
then x = 222 ÷tan 45
& y = 222 ÷ tan 60
distance between the ships = x + y = 350.172 m

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Auditya   7 years AGO

Let A and C are ships and B is light house of 222 m height
Angle of elevation to B from A and C are 45 and 60 degrees respectively
Tan=opp side / adj side
Tan 45 = 222/AB ~1=222/AB
~AB=222 m
We have √3=1.732
Tan 60=222/BC~√3=222/BC
~BC=222/√3=128.171m
AC distance =AB+BC=222+128.171=350.171m

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ribca   7 years AGO

distance of ship 1 = x = 222/(tan 45) = 222 m
distance of ship 2 = y = 222/(tan 60) = 128.175 m

distance between ships = x + y = 222 + 128.175 = 350.17 m

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parul mehta   7 years AGO

Applying PGT to both sides of altitude(lighthouse) finding bases of both triangles and then adding them

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