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suppose two people are standing on a beach 50 meters apart. Each person looks at an island off in the distance, one has a 30 degree line of sight and the other has a 40 degree line of sight. How far is it from the beach to the island (find the shortest, straight line distance).

Guest Nov 19, 2018
 #1
avatar+92787 
+1

The distance from the beach to the island  will form the altitude, A, of a triangle

Let x represent the distance  that the person with the 30 degree line of sight is from where the altitude intersects the base

So...the distance that the person with the 40 degree line of sight to this intersection point is 50 -x

 

So   we have this

 

tan 30  = A/ x       

 

Rearrange as  [ x  =  A / tan 30  ] ⇒

 

[ x =    A / (1/√3)  =  √3A ]

 

And we have that

 

tan 40 = A / (50 - x)       sub in for x  and we get that

 

tan 40  = A / ( 50 - √3A)

 

(50 - √3A) * tan 40  = A

 

50sin 40 - √3A tan40  = A

 

50tan 40   =   √3Atan 40 + A          factor out A on the right

 

50tan 40  =  A ( √3 tan 40 + 1)      divide both sides by  (sin 40 + 1)

 

A = 50tan 40 / [√3 tan 40 + 1 ] ≈   17.1 (m)

 

EDIT TO CORRECT AN ERROR!!!

 

 

cool cool cool

CPhill  Nov 19, 2018
edited by CPhill  Nov 19, 2018
edited by CPhill  Nov 19, 2018
 #2
avatar+14578 
+1

Slightly different approach using law of sines

 

x/sin30 = 50/sin10   yields x = 26.6044 (distance from 40 degree observer to island)

 

then 

 

26.044/sin 90 = x /sin40    yields   x = 17.1 m shortest distance to the beach....

ElectricPavlov  Nov 19, 2018
edited by ElectricPavlov  Nov 19, 2018
edited by ElectricPavlov  Nov 19, 2018

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