A periscope is 5 feet above the surface of the ocean. Through it can be seen a ship that rises to 50 feet above the water. To the nearest mile, the farthest away that the ship could be is ( ) miles.
Well I found the answer here, through Google rabbi. It's a question required a little imagination haha https://answers.yahoo.com/question/index?qid=20140722052535AARfCNZ
quoting the previous reply: 2. Sketch this, assuming the Earth is a sphere with radius R, so a plane slice through the periscope, ship and center of the Earth is a circle of radius R. Draw the lines out from the center (O) of the circle to point A at the top of the periscope and point B at the top of the ship. so that line AB is tangent to the circle at point C. That makes triangles OAC and OBC right triangles, each having the right angle at C. From the problem, the lengths OA = R+5, OC = R, and OB=R+50. Label the lengths AC = p and BC= q, then use Pythagoras: R² + p² = (R + 5)² R² + q² = (R + 50)² Solve those: p² = (R + 5)² - R² = 10R - 25 p = √(10R + 25) q² = (R + 50)² - R² = 100R + 2500 q = √(100R + 2500)
Can you help me?
It's confusing
@amistre64
the first responder is suggesting that the curvature of the earth play into effect; which i have no idea if that is applicable to this question, it may be tho
|dw:1406569521046:dw|
I dont think this is applicable to this question
then if we are dealing with a flatworld interpretation, then the distance one thing can be from another approaches infinity. is field of depth involved? what dort of context can you provide?
? I'm not sure. A periscope is 5 feet above the surface of the ocean. Through it can be seen a ship that rises to 50 feet above the water. To the nearest mile, the farthest away that the ship could be is blank miles
i think the circular stuff is applicable
ok
think of a circle that has a radius of (earth+5) in feet and a tangent line that intersects the diameter line at (earth+50) feet. then the arclength between them is the distance |dw:1406569844637:dw|
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