Math Is Fun Forum

I was wondering, how could I find the exact value of arctan(−10) ? We know that an approximation of the exact value would be arctan(−10)≈−1.47112 rad but if we wanted the exact value in radians, How would I find it ?

Thank you!

Hi, if we have a scalene triangle with : 112,28 and 40 degrees. and that we have a side of lenght 4 between 28 and 40. Can we find the height of the triangle without using the sine law ?

Thank you!

Hi, there's just one thing which I'm not sure with this problem :

There are 24 pounds of nails in a sack. Can you measure out 9 pounds
of nails using only a balance with two pans?

I answered yes and here's the reason. Consider a number X of nails which weights 24 pounds. This number X has atleast a factor of 2^3 (It can be higher.) in its factors. Why ? Because we need to divide evenly the number of nails so we can have equal weight on each side of balance.

So :
x/2=x/2

x/4=x/4

x/8=x/8

x/2=12 pounds x/4=6 pounds x/8=3 pounds

So we have the following groups (after successive divisons.):

x/2+x/4+x/8+x/8=x

Clearly, x/4+x/8 nails would give 9 pounds

Now, the only thing bugging me is that my solution is founded on the assumption that all nails must have equal weights. The question isn't clear if the nails do have all the same weight or not...

What do you think ???

THank you!

All the proportions which we have deduced from a:b::c:d may be represented generally as follows :

ma+nb:pa+qb::mc+nd: pc + qd

For the product of the extreme terms is mpac+npbc+mqad +nqbd; which, since ad = bc, becomes mpac+npbc+mqad +nqbd; also the product of the mean terms is mpac+mqbc+npad+nqbd; or, since ad = bc, it is mpac+npbc+mqad +nqbd: so that the two products are equal.

Here are some of the proportions deduced from a:b::c:d   :

(a-b):a::(c-d):c
(a+b):a::(c+d):c

and some others (I'm not putting all of them.)

I don't understand the part : "All the proportions which we have deduced from a:b::c:d may be represented generally as follows :

ma+nb:pa+qb::mc+nd: pc + qd"

What does it mean ??? I'm not following for a certain reason....

Thank you!