isosceles and tight triangle

Question

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Hi, I was wondering if there is a way to find x with only knowing the length of isosceles triangle and no other piece of information.

You mean only knowing then length of $OB$? If so, then no, you can't. However, if you know $AB$ or the angle $A,B$ or $O$, as well, then you can. — BeaumontTaz, Aug 12 '14 at 16:39
I think your gonna need more information. As stated, the angle At O can be anything without changing the setup — Mathmo123, Aug 12 '14 at 16:40
It probably looks like $\angle ABC$ is $30^\circ$, taking that as an assumption we can calculate x.Otherwise, you can't. — MonK, Aug 12 '14 at 16:57

score 0 · Answer 1 · answered Aug 12 '14 at 16:47

The answer is no. You cannot determine it solely on $\overline{OB}$. Because the length of $\overline{AB}$ isn't defined. You can make a really fat isosceles triangle by making $\overline{AB}$ large or you can make a really skinny isosceles triangle by making $\overline{AB}$ small. This would change $x$.

If you fixed any angle in the isosceles triangle then you can solve it. Or if you fix the length $\overline{AB}$ then you can solve it.

score 0 · Answer 2 · answered Aug 12 '14 at 16:58

If $\angle AOD=\angle DOB=y^\circ,\angle OBA=\angle OBD=(90-y)^\circ\implies \angle ABC=y^\circ$

For the right angular triangle $\displaystyle\triangle ABC,\frac x{AB}=\sin y^\circ\implies x=2BD\sin y^\circ$

For the right angular triangle $\displaystyle\triangle ODB,\frac{BD}{650}=\sin y^\circ$

$\implies x=2\sin y^\circ(BD)=2\sin y^\circ(650)\sin y^\circ=1300\cdot\sin^2y^\circ$

Clearly, the value of $y$ can be arbitrary between $(0,90)$

isosceles and tight triangle

2 Answers2