A clock has a minute hand 6 cm in length, and an hour hand of 4.5 cm. Assuming the clock keeps perfect time, how far apart are the hands (at their tips) at twenty-five past two (2:25)?
This was in a recent maths test of mine. At first I thought the hands formed a simple right angle, but that can't be the case, given that the hour hand won't be directly over the '2'. What say ye, math folk?
You have the diameters of the circles, the radius are the sides of the triangle. The time swept is little hand on the 2, big hand on the 25. That should be 90 degrees. Use the angle and the 2 radii to calculate the last side.
Giant Purple Snorklewacker wrote: You have the diameters of the circles, the radius are the sides of the triangle. The time swept is little hand on the 2, big hand on the 25. That should be 90 degrees. Use the angle and the 2 radii to calculate the last side.
But if the clock keeps perfect time, at 2:25, the hour hand will be 25/60 of the way between the 2 and the 3.
Big hand on the 25?
I think you mean Little hand on the 2 and big hand on the 5.
Which is 3 apart, which is 30 degrees, which would be 90 degree total (360 degrees divided by the number of numbers on a clock face (12) = 30 degrees per number.)
(Just wanted to clarify, you made a few assumptions that needed to be shown.)
Given they want the straight line distance and not the arc you have the Pythagorean theorem. A^2+B^2=C^2. Solve for C
1988RedT2 wrote:Giant Purple Snorklewacker wrote: You have the diameters of the circles, the radius are the sides of the triangle. The time swept is little hand on the 2, big hand on the 25. That should be 90 degrees. Use the angle and the 2 radii to calculate the last side.But if the clock keeps perfect time, at 2:25, the hour hand will be 25/60 of the way between the 2 and the 3.
Clarification: By perfect time we have to make the assumption that it is not a mechanical clock. The hour hand stays on the 2 until exactly 3. Which is the 10 minute mark. The minute hand points at the 25. 25 - 10 = 15. 15 minutes (of time...) is a 90 degree angle.
It would be unreasonable to assume that the problem author meant for you to take in to account the hour hand's "creep" induced by the mechanism without some mention of the rate at which that happens.
FlightService wrote: Big hand on the 25? I think you mean Little hand on the 2 and big hand on the 5. Which is 3 apart, which is 30 degrees, which would be 90 degree total (360 degrees divided by the number of numbers on a clock face (12) = 30 degrees per number.) (Just wanted to clarify, you made a few assumptions that needed to be shown.) Given they want the straight line distance and not the arc you have the Pythagorean theorem. A^2+B^2=C^2. Solve for C
I didn't see this until after ... but yeah, I could have been clearer the 1st time. For extra credit he should calculate the arc since you gave him the formula for this one ;)
In reply to Giant Purple Snorklewacker: Right so if it is a true mechanical clock it would be 90 degree - (25/60 * 10 degrees).
which would be 85.833 degrees.
So you could use the definition of sin (or cos) so sin (85.833 degrees)=...
You should be able to get the rest
If we consider the circle of the clock face, which of course has 360 degs., we can put the minute hand at 150 and the hour hand (assuming the aforementioned "creep") at 72.5 deg. The rate of "creep" of any mechanical clock can be assumed to be regular, i.e. at 25 past the hour, the hour hand will have travelled 26/60 of the way to the next number.
To find the included angle, we subtract 72.5 from 150 yielding 77.5 degrees between the hands. Using the cosine rule, the answer of 6.67 cm is elementary. 
FlightService wrote: In reply to Giant Purple Snorklewacker: Right so if it is a true mechanical clock it would be 90 degree - (25/60 * 10 degrees). which would be 85.833 degrees. So you could use the definition of sin (or cos) so sin (85.833 degrees)=... You should be able to get the rest
Actually, you really don't need to do that. 15 minutes of time is a true 90 degree angle. All you need are the radii values from there.
it's a 3-4-5 triangle with a base of 1.5
1.5 x 5 = 7.5
Giant Purple Snorklewacker wrote: Clarification: By perfect time we have to make the assumption that it is not a mechanical clock. The hour hand stays on the 2 until exactly 3. Which is the 10 minute mark. The minute hand points at the 25. 25 - 10 = 15. 15 minutes (of time...) is a 90 degree angle. It would be unreasonable to assume that the problem author meant for you to take in to account the hour hand's "creep" induced by the mechanism without some mention of the rate at which that happens.
That's the answer I went with. Looking back on it, I figured there must've been more to the problem. Looks like I was just over-thinking.
edit: or not! 1988RedT2 is correct I believe.
Yup, 6.67 is right.
I did it by just plotting the location of the end points and using the distance formula d=sqrt { (x1-x2)^2 + (y1-y2)^2 }
so hour hand is 72.5 degrees from vertical, so it's end points is at (4.29, 1.353). Angle comes from 360/12 for each hr, so 2 o'clock on the dot is 60 degrees. The creep comes from 30degrees/60 minutes, so each minute is .5 degrees, so the end angle is 2(30) + .5(25)=72.5
Minutes are easier, 360/60=6 degrees per minute. 6(25)=150 degrees. Use sine and cosine to find end points of (3, -5.196). Use distance formula to calculate distance.
I suppose, the more I think about it... yeah, you should have calculated the linear sweep of the hour hand. I have little kids - geometry homework is typically: find the shape and calculate the missing part. Since you are probably not in middle school I suppose I should have given the test author a little more credit.
In the future... I'll still over simplify all problems presented. Apply accordingly ;)
This is why they invented digital clocks.
Anyone have the Far Side cartoon of Hell's Library?
And now we know why engineering is done by commitee
In reply to Giant Purple Snorklewacker:
Yeah you do, it works even better when you follow your own description and realize each section is 30 degrees and not 10
90-(25/60*30) = 77.5 degrees
And I wonder why I never got a 100 in math.
Miata.
I was gonna say half past the monkeys balls.
Of all the various attributes of the people who frequent this forum, I do enjoy the sophistication the most.
Assuming the clock is real I would have just measured it. Assuming it isn't, who cares.
Problems like this were why I hated math. If I'm going to figure out a problem at least make the answer worth knowing.
And this is why I didn't become an engineer folks.
EastCoastMojo wrote: I was gonna say half past the monkeys balls.
Not much makes me laugh out loud. LOL.
Toyman01 said: Problems like this were why I hated math. If I'm going to figure out a problem at least make the answer worth knowing.
I never hated math, just the teachers. Regardless your point is well taken. What's the point? Real world applications work for me and my students. The same concepts could be figured building a set of stairs. Use brains and hands, get results.
I get the whole "make em really think part" but seriously. BTW, what grade level we talking about here, HS or College?
Without checking Brotus' work, that sounds right. Digital clocks? Not the point, the point is to find the equasion is not the answer.
Much like a problem I had as an aspiring machinist; there's a fly here, you are there, both inside a cube. Find the shortest path to the fly.
A test of how you resolve problems?
Dan
vwcorvette wrote: BTW, what grade level we talking about here, HS or College?
1st year University. As a Biology major, this is the one and only maths class I take, (apart from some Statistics, later on.)
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