In a quadrilateral $ABCD$ are known $\overline{AB}=6a'\sqrt3$, $\overline{AD}=15a'\sqrt2$ and the angles $D\hat{C}A=\pi/4$, $A\hat{B}C=2\pi/3$, $A\hat{C}B=\pi/6$. Calculate the measure of the diagonal $AC$ and the perimeter of the quadrilateral.
The solutions of my textbook are: $\overline{AC}=18a'$ and the perimeter is $2p=(12\sqrt 3+36\sqrt 2)a'$.
I put the original Italian question:
My synthetic solution: I have drawn the image:
We have:
$$\frac{b}{\sin \beta}=\frac{c}{\sin \gamma} \implies b=18 a', \quad \frac{a}{\sin \alpha}=\frac{b}{\sin \beta} \implies a=6a'\sqrt 3$$
Hence $c=a=6a'\sqrt 3$. I know that $\eta=\pi/3$ and $\delta=\pi/4$. Thus $\psi=5\pi/12=15°$. Now
$$\frac{15a'\sqrt 2}{\sin \delta}=\frac{\overline{DC}}{\sin \psi} \implies \overline{DC}=30a'\cdot \sin(15°)$$ and $$\sin(15°)=\sqrt{\frac{1-\cos(\pi/6)}{2}}=\frac{\sqrt 6 -\sqrt 2}{4}$$
Possibly I will have made some mistake or there is an error in the textbook of an high school but I will never find that perimeter of the solution.
Lastly, my female student has forwarded me her solution
and she asks me how her could derive from the arcosine $\text{arcsin}=\sin^{-1}$ a non-approximate relation to derive the correct misure of the angle when it is:
$$\sin^{-1}\left(\frac 35\right)$$
(see the red rectangles).
Here is your mistake: $\eta \ne \pi/3$
$\displaystyle \frac{AD}{\sin \delta} = \frac{b}{\sin \eta} \implies \sin \eta = \frac{3}{5}$ and so $\displaystyle \cos \eta = \frac{4}{5}$
Now, $\displaystyle \sin \psi = \sin (\pi - ({\eta + \frac{\pi}{4}})) = \sin (\eta + \frac{\pi}{4}) = \frac{1}{\sqrt2} (\sin \eta + \cos \eta) = \frac{7}{5 \sqrt2}$
So using $\displaystyle \frac{CD}{\sin \psi} = \frac{AD}{\sin \delta}$, $CD = 21 \sqrt2 a'$.
As you know all sides, adding them leads to perimeter of $(12 \sqrt3 + 36 \sqrt2) a'$.