If

are three angles of a

such that

and

Then all possible values of

Source :

i have tried like using

So

So

are roots of the equation

for real roots ,

So

or

but answer is

Thanks

Hello!

It is possible to post the whole problem in one block of latex.
(Just click that "cite" button [Citeaza] to see how i retyped the text.)

In the following i will mark one passage in red.

[Citat] Thanks

Yes, it is all ok, but at one important point there is no equivalence.

So from the given data, p is constrained to be in the set established above.
But is any such value possible?

There is a simple argument to show, that the values from zero (well, zero itself... is first a question about the definition of a "triangle" - but it should not be degenerated, if we already know an angle...) to 3 - 2sqrt(2) > 0 cannot be reached! It goes like this: Let us say, that such a number p is realized. It is then realized for two roots having the same sign. The sum has also this common sign. It is the sign of p-1. OK, negative roots.
But this is a strange triangle, since we expect two angles between 0° and 90° in it...

A more straightforward way to proceed, also using some of the above ideas is as follows.

Let x be tan(B).
And a first important question is: Which are the possible values of x? The possible values and only the possible ones. (If we miss this point, we miss the point.)

Then write tan(C) = tan(135°-B) = ... as a function of x = tan(B) .
Then consider the product p as a function of x.

And use the prozaic study of this function.
(It has a plot, so one is always doing the things "right", with a possibility of control... Doing so, one always gets "some points" in an exam. Because the progress is vizible - the given problem is reduced to a standard one.)

One www.wolframalpha.com one can try:
plot x*(x+1)/(x-1)
plot x*(x+1)/(x-1) from 0 to infinity
plot x*(x+1)/(x-1) from -infinity to -1
max x*(x+1)/(x-1)
min x*(x+1)/(x-1)

Thanks Admin.