The captaian has an order to go from base A to port B
in shortest possible route. He sets the ship course to NW
and leaves the base. When en route, another order arrives
redirecting the ship to port C. The captain turns the ship
90 degrees to the left. After some time a fire abroad the ship
interrups her voyage. The captian makes emergency
turn 130 degrees to the left and proceeds back to the base A.
The ships retuns to her base from the west and enters repair
dock. What is the area of the sea enclosed in the loop?
2007-02-01 06:10:59 · 4 answers · asked by Alexander 6 in Science & Mathematics ➔ Mathematics
Good thing Gordon B at last noticed that
it's not S = 1/2 ab sin C problem.
2007-02-01 08:44:45 · update #1
The Pacific Ocean, South China sea
2007-02-01 06:19:02 · answer #1 · answered by whatevit 5 · 0⤊ 0⤋
Chaney is right in his analysis of the problem but there sre two flaws in the problem that make it incalculable. The resultant triangle has 3 inside angles 90 degrees, 50 degrees and 40 degrees.
To directly return to base "A" from the West (a heading of 90 degrees) is impossible in this diagram (given true headings.) If you turn 130 degrees along the line BC to return directly to A your resultant heading is 95 degrees (or not true East,) You said FROM the West.
Given a relationship of 3,4,5 for the right angled triangle, the area is a value of 6 (if the hypotenuse is 5) It could be 0.6, 6, 60 or 600 square miles dependent on the actual length of one side which is not defined.
The only possible explanation would be the difference in magnetic variation somewhere in the world which would make the last turn of 130 degrees result in an actual (or magnetic) heading of 90 degrees (-5 degrees variation) but I don't have an isobaric chart with lines of magnetic variance at hand. That would possibly also allow you to calculate the length of one side...
Good luck!
2007-02-01 08:31:31 · answer #2 · answered by Gordon B 4 · 0⤊ 0⤋
The area calculated is for a right triangle. So A =1/2*base*height.
Starting point is A, then to B(90degree turn to left), then to C(130 degree turn to left returning to A).
Base is segment BC, Height is segment AB. Hypotenuse is segment AC.
Angle at B is 90. Angle at C is 50 (180-130=50). Angle at A is 40(180-90-50=40)
2007-02-01 06:24:47 · answer #3 · answered by Anonymous · 0⤊ 0⤋
You would need at least one distance. There are no distances given in your problem.
2007-02-01 06:41:09 · answer #4 · answered by Surveyor 5 · 0⤊ 0⤋