When you made that DE line... How is that guaranteed to be the same as CD? (Edit) I see, the given 45° makes a 45°/45°/90° triangle. I missed seeing that at first.
Kept on thinking I had solved it, etc...I think I got it now, maybe there is a better solution... I started thinking yesterday that I could think of it as a increasing AD, etc...like A and C are fixed, and B is changing position following the length of AD, etc...DB could could be seen as a projection of some light source, lol, or something...AC could be on the y-axis of a xy-plane, etc, and AB on the x-axis, etc...but I don't think any of that nonsense is really necessary, I was thinking of that fact that, clearly, the angle 45 would change depending on the length of AD, etc, like it's a function of AD...knowing one means knowing the other, etc, and either determines the hypotenuse, etc, the position of the point D fixes everything else, if A and C are fixed, etc...basically I was imagining movement, a changing triangle, etc, increasing AD, but I don't think any of that is necessary... So, lol, 45 degrees = pi/4 (if one wants) = arctan [(5+AD)/3 ]- arctan(AD/3), etc (I mean, lol, if tan=opposite/adjacent, lol, I forgot a lot of these basic things...and don't ask me about the arctangent function, lol, it will obviously yield unique results within certain ranges, etc, I forgot how one restricts the arctangent, arcsin, etc...I used the arctangent because we "have" the opposite and adjacent)...set pi/4 equal to all that nonsense, find what AD is...the moment one has AD, lol, one can easily use Pythagoras to get the hypotenuse...it's so simple...no need for any sliding, but it's the thinking about the sliding, the correspondence between the angles and the length of AD, etc, that made me realize that, etc, while trying to find what the dependency...overall, should have taken me 5 mins, lol, don't know why it took me as long as it did...hopefully I didn't screw again and haven't solved it...
When you made that DE line...
How is that guaranteed to be the same as CD?
(Edit) I see, the given 45° makes a 45°/45°/90° triangle. I missed seeing that at first.
But how can they be equally long
Playing with tangents calculate length of AD
then from Pythagoras' theorem calculate length of BC
I solve it like your way
Kept on thinking I had solved it, etc...I think I got it now, maybe there is a better solution...
I started thinking yesterday that I could think of it as a increasing AD, etc...like A and C are fixed, and B is changing position following the length of AD, etc...DB could could be seen as a projection of some light source, lol, or something...AC could be on the y-axis of a xy-plane, etc, and AB on the x-axis, etc...but I don't think any of that nonsense is really necessary, I was thinking of that fact that, clearly, the angle 45 would change depending on the length of AD, etc, like it's a function of AD...knowing one means knowing the other, etc, and either determines the hypotenuse, etc, the position of the point D fixes everything else, if A and C are fixed, etc...basically I was imagining movement, a changing triangle, etc, increasing AD, but I don't think any of that is necessary...
So, lol, 45 degrees = pi/4 (if one wants) = arctan [(5+AD)/3 ]- arctan(AD/3), etc (I mean, lol, if tan=opposite/adjacent, lol, I forgot a lot of these basic things...and don't ask me about the arctangent function, lol, it will obviously yield unique results within certain ranges, etc, I forgot how one restricts the arctangent, arcsin, etc...I used the arctangent because we "have" the opposite and adjacent)...set pi/4 equal to all that nonsense, find what AD is...the moment one has AD, lol, one can easily use Pythagoras to get the hypotenuse...it's so simple...no need for any sliding, but it's the thinking about the sliding, the correspondence between the angles and the length of AD, etc, that made me realize that, etc, while trying to find what the dependency...overall, should have taken me 5 mins, lol, don't know why it took me as long as it did...hopefully I didn't screw again and haven't solved it...
What i did was that took angle C to be 45+Q and solved for tan after that finded out the value of x...
In my opinion this is the easiest solution
I did the same
And just how did you go about finding the value of the tangent of angle C ?
whats wrong with pythagorus? . Much simpler
Are you saying that you can just use Pythagoras directly on △ABC?
Show us.
Actually, no, the length of the hypotenuse is sqrt (58).
I got the sq root of hypotenuse be : 6.7082039
Perhaps you should show your calculations here for all to see .
Show your working.