- #1
georg gill
- 153
- 6
one want to find the length of the two hypotenuses of the figure
Here is what I did. I looked at the small first with hypotenus unknown and adjacent=1 and the other cathetus unknown. Then i wanted to find max value of the smallest hypotenus.
[tex]tan\theta=\frac{1}{a}[/tex][tex]a=\frac{1}{tan\theta}[/tex] Smallest hypotenus=sh
[tex]sh^2=1+\frac{1}{tan^2\theta}[/tex]
i took derivative to find max value
[tex]sh=\sqrt{1+cot^2\theta}[/tex]
[tex]\frac{d}{dx}sh=\frac{1}{\sqrt{1+cot^2\theta}} \cdot \frac{coth\theta}{sin^2\theta}[/tex]
but this is equal to zero when [tex]\theta=\frac{\pi}{2}[/tex] whoch gives sh=1which is obviously wrong since adjacent cathetus=1
What is wrong?
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