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# The twelve hints of hercules

(unpublished)

Spoiler Alert: This post gives a lot of hints on how to do the problem about the famous bullwhip that I posted on Dec 3, 2012 under the heading “Sine of the times”. If you got stuck on the problem, read a hint and then give it another try. Only read another hint if you are still stuck. You might want to have someone read you the hints so that you do not accidentally read more hints than you need.

Hint #1: Since the problem did not provide a diagram, and there are a lot of confusing details, read through the problem carefully and draw a diagram.

Hint#2: The diagram should include the known quantities and some indication of the unknown quantity (the angle).

Hint#3. The diagram should show that this problem boils down to the solving of a triangle.

Hint#4: The type of triangle being solved is SSA.

Hint#5. In particular, the triangle being solved is a sSA triangle, which means that the side next to the given angle is bigger than the other provided side.

Hint#6. There is a special consideration you must be aware of when solving sSA triangles.

Hint#7: The special consideration has to do with there being more than one possible triangle that can be considered a valid solution when given the inputs in the sSA form.

Hint#8: The triangle being solved is a sSA triangle, where the bottom of the triangle (pointing east from the original location of the priceless artifact, in the direction of our travel, is 20 yards, or 60 feet). The other known side length is the length of the bullwhip plus the length of your arm (42 feet). The provided angle is from the original location of the priceless artifact, an angle of 30 degrees north of our direction of triangle (60 degrees east of north, or in other words, clockwise from ninety degrees to the left of our direction of our escape route).

Hint#9: To find one solution of a sSA triangle, use the sine law to solve for the angle between the 42 foot outstretched bullwhip/arm and the unknown length of laser between its source and the place where it will be interrupted.

Hint#10: If you find the simplest solution to the sine law, and if you have read the problem carefully, you will find that the acute angle you have solved for results in a triangle which indicates that you should project the bullwhip obstruction in a direction that the problem stated was not allowed, due to the obstructions from falling debris in that direction. If you showed this restriction in your diagram, you probably remembered this fact and spotted this conflict.

Hint#10: To find the other solution, realize that for any angle theta satisfying the relation: sine theta equals x, there are two angles theta that satisfy the equation, unless x = 1. The first angle solution is the arcsine of x, and the other is 180 degrees (or pi, if using radians) minus the first solution.

Hint#11: You should choose the second solution, so that your triangle solution will give you an answer for your angle of “throw” that does not get blocked by falling debris.

Hint#12: The angle you need to solve for can be found by using the rule about what the sum of the interior angles of a triangle must always add up to.

If you still need more hints, let me know. Or if you simply enjoyed solving this problem, please let me know that too, and I will gladly generate more adventure problems like them. Let me know if there is a particular topic of math, physics, or chemistry you would like to be challenged on or challenge your friends with.

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Isaak B.

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