Maximizing the cross-sectional area of an isosceles trapezoid with an angle z I have been given the following problem:

A canal must be excavated, which must have a cross section shaped like an isosceles trapezoid. In the cross section, the
bottom line and the two slanting side pieces below the waterline must
together have a length of 150 m.
Determine the angle z so that the water flow can be as large as possible, ie. that the cross-sectional area must be as large as
possible

Image of the isosceles trapezoid
My drawing of the figure with defined sides
First of all, I have divided the trapezoid into one rectangle and two triangles.
In regards to the length of the sides I have said that:
$$150=120+2c\Leftrightarrow 30=2c\Leftrightarrow 15=c$$
I have determined a and b to be:
$$sin(z)=\frac{a}{c} \Leftrightarrow sin(z)\cdot c=a$$
$$cos(z)=\frac{b}{c} \Leftrightarrow cos(z)\cdot c=b$$
so that I can substitute these into the formula for the area of a triangle:
$$A_{triangle}=\frac{1}{2}\cdot a\cdot b=\frac{1}{2} \cdot sin(z) \cdot c \cdot cos(z)\cdot c=\frac{1}{2}c^2 \cdot sin(z) \cdot cos(z)$$
In the same way I have defined the area of the rectangle as:
$$A_{rectangle}=a \cdot 120 = sin(z) \cdot c \cdot 120$$
When I combine the area of the rectangle and the area of the triangle to get the area of the trapezoid I get:
$$A_{trapezoid}=2 \cdot A_{triangle} + A_{rectangle} =\frac{1}{2}c^2 \cdot sin(z) \cdot cos(z)+sin(z) \cdot c \cdot 120$$
Then I find the derivative of the area of the trapezoid:
$$A_{\text {trapezoid }}^{\prime}=\frac{d}{d z} A_{\text {trapezoid }}(z)=\frac{d}{d z}\left(c^{2} \cdot \sin (z) \cdot \cos (z)\right)+\frac{d}{d z}(\sin (z) \cdot c \cdot 120)$$
$$\quad=c^{2} \frac{d}{d z}(\sin (z) \cdot \cos (z))+120 \cdot c \frac{d}{d z}(\sin (z))$$
$$\quad=c^{2} \cdot \cos (z) \cdot \cos (z)+\sin (z) \cdot(-\sin (z))+120 \cdot c \cdot \cos (z)$$
$$\quad=c^{2} \cdot \cos ^{2}(z)-c^{2} \cdot \sin ^{2}(z)+120 \cdot c \cdot \cos (z)$$
Then I let: $cos^2(z)-sin^2(z)=cos(2z)$
At last I try my best to solve for z, but this is as far as I got:
$$\begin{aligned} c^{2} \cdot \cos ^{2}(z)-c^{2} \cdot \sin ^{2}(z)+120 \cdot c \cdot \cos (z)=0 & \Leftrightarrow\end{aligned}$$
$$\begin{aligned} c^{2}\left(\cos ^{2}(z)-\sin ^{2}(z)\right)+120 \cdot c \cdot \cos (z)=0 \Leftrightarrow \end{aligned}$$
$$\begin{aligned}c^{2} \cdot \cos (2 z)+120 \cdot c \cdot \cos (z) &=0 \Leftrightarrow\end{aligned}$$
$$\begin{aligned} c^{2} \cdot \cos (2 z)+120 \cdot c \cdot \cos (z)=0\end{aligned}$$
However, when I try to solve for z using a CAS-tool then $z = 83.03169^\circ$. Although, the answer in my textbook says $z = 96.97^\circ$.
Why is this and what have I done wrong?
Thanks a lot in advance for the help! :)

There is another way to do this using the pythagorean theorem. The area of a Trapezoid is half the sum of the bases times the perpendicular height. If we assign $x$ to the $b$ dimension in your diagram then:
$$A = (120+x)(15^2-x^2)^{\frac{1}{2}}$$
Maximum area occurs where: $$\frac{dA}{dx} = 0$$
$$(15^2-x^2)^{\frac{1}{2}}+\frac{-2x(120+x)}{2(15^2-x^2)^{\frac{1}{2}}}=0$$
$$(15^2-x^2)^{\frac{1}{2}}=\frac{x(120+x)}{(15^2-x^2)^{\frac{1}{2}}}$$
$$15^2 - x^2 = 120x + x^2$$
$$2x^2+120x -225 = 0$$
Using the quadratic formula: $$x = 1.81981$$
The supplementary angle to angle $z$ is: $$\cos^{-1}(\frac{1.81981}{15}) = 83.03^o$$
$$\text{Angle}\ z  = 96.97^o$$
