# Finding the height and base of an isosceles triangle whose equal sides are given.

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?

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

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 $h=\sqrt{c^2-\left(\frac{a}{2}\right)^2}$