Find one side of a triangle using trigonometry impossible question In the isosceles triangle $ABC$ where $a=|BC|,\ b=|CA|,\ c=|AB|$ and angle at corner $A = x$
and angle at corner $B = y$ and angle at corner $C = z$ ($z>90$ degrees)
$a=b=4$ ; $\sin z = 3/8$
Find value of $c$ (which is side $|AB|$ as stated above) 
My friends father is math teacher in high school and he couldn't even solve it,
can anyone help?
 A: Simply use law of cosine:
$$c = \sqrt{a^2 + b^2 - 2ab\cos z} = \sqrt{a^2 + b^2 - 2ab(-\sqrt{1-\sin^2 z})} = \sqrt{a^2 + b^2 + 2ab\sqrt{1-\sin^2 z}}.$$
EDIT: Because $z> \pi/2$, $\cos(z) <0 \implies \cos(z) = -\sqrt{1-\sin^2z}$
A: @grixor's answer, using the law of cosines is (currently, almost) correct, but perhaps uses a tool that is not yet in the student's toolbox.  We can derive this result from the problem as given.
Extend $\overline{AC}$ until it intersects the perpendicular to $\overline{AC}$ through $B$ and call this intersection $D$.  Call the measure of angle $BCD$ $\theta$ and note that $\theta = 180^\circ - z$.  Then the right triangle $BCD$ supplies \begin{align}
    \sin \theta &= \frac{|BD|}{a}  \\
    \cos \theta &= \frac{|CD|}{a}  \text{.}  \\
\end{align}
Using supplementary angle identities, we have $\sin \theta = \sin z = |BD|/a$ and $\cos \theta = -\cos z = |CD|/a$.  Rewriting in terms of the quantities we want, $|BD| = a \sin z$ and $|CD| = -a \cos z$.
Now shift attention to the right triangle $ABD$.  From this triangle and the Pythagorean theorem, we see \begin{align}
    c^2 &= |BD|^2 + (b+|CD|)^2  \\
        &= (a \sin z)^2 + (b - a \cos z)^2  \\
        &= a^2 \sin^2 z + b^2 - 2 a b \cos z + a^2 \cos^2 z  \\
        &= a^2( \sin^2 z + \cos^2 z) +b^2 - 2 a b \cos z  \\
        &= a^2 + b^2 - 2 a b \cos z  \text{.}
\end{align}  We have used the identity $\sin^2 z + \cos^2 z = 1$.  We use it again to get $\cos z = -\sqrt{1 - \sin^2 z}$.  (We know we want the negative root because we know $z$ is in the second quadrant, so cosine is negative.)  Collecting our algebra together, we have  $$  c^2 = a^2 + b^2 + 2 a b \sqrt{1 - \sin^2 z}  \text{.}  $$
We're given $a = b = 4$ and $\sin z = 3/8$, so we can evaluate:\begin{align}
    c^2 &= 4^2 + 4^2 + 2 \cdot 4^2 \sqrt{1 - (3/8)^2}  \\
        &= 32 + 32\sqrt{\frac{64-9}{64}}  \\
        &= 32 + 32\sqrt{55/64}  \text{.}
\end{align}  Now we know $c = \sqrt{32 + 32\sqrt{55/64}}$.
