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Let the given side be $5$;

My approach to solving this is:

let $\phi$ be the angle formed by the side and height of the triangle, and $θ$ be the angle between the base and side.

then,

$$2θ + 2\phi = 180$$

or,

$$ \sin^{−1}(h/5) + \sin^{−1}\Big(\frac{b/2}{h}\Big) = \sin^{−1}(1)$$

or,

$$h/5 + b/2h = 1 $$

or,

$$2h^2 + 5b = 10h \hspace{1cm} \text{---- eqn (1)}$$

Using Pythagorean theorem:

$$h^2 + (b/2)^2 = 5^2 \hspace{1cm} \text{---- eqn (2)}$$

Now, solving these two equations and getting h and b doesn't seem good. So how should I find out the $h$ and $b$ when I am given with one side of the isosceles triangle?

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  • $\begingroup$ No angles given? $\endgroup$ – QuIcKmAtHs Dec 31 '17 at 4:11
  • $\begingroup$ @XcoderX nope... $\endgroup$ – Amit Upadhyay Dec 31 '17 at 4:13
  • $\begingroup$ The angle between the 2 equal sides is not determined, it is unsolvable $\endgroup$ – QuIcKmAtHs Dec 31 '17 at 4:14
  • $\begingroup$ $\sin^{-1}(90)=1$?? What's that? $\endgroup$ – velut luna Dec 31 '17 at 4:14
  • $\begingroup$ We have to address the problem. It provides insufficient info $\endgroup$ – QuIcKmAtHs Dec 31 '17 at 4:15
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You can't solve it based only on the side given. The problem is missing information.

If the side is $c$, pick any length $a$, $0\lt a\lt 2c$ - and you can construct an isosceles triangle with base $a$ and side $c$.

Once the base is fixed, the triangle is uniquely determined up to congruency, and the height corresponding to the base is $h=\sqrt{c^2-\left(\frac{a}{2}\right)^2}$

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Having a side of an isosceles triangle is not sufficient to construct such a triangle. The base length could be any number greater than zero and less than twice the length of the given side.

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