Tomek and Piotr solved the following task:
In a right-angled triangle $ABC$, $|BC|=3$, and $|AC|=4$. On the hypotenuse $AB$ there is a point $D$, and the measure of the angle $ACD$ is $60^{\circ}$. Calculate the length of the line segment $CD$.
Tomek presented this solution:
(1) He sketched a picture of a right-angled triangle.
(2) He expressed the area of the triangle $ABC$ as a function of $x$, where $x=|CD|$: $$ \begin{gather} P_{ABC}=P_{ACD}+P_{BCD} \cr \frac12 \cdot 3 \cdot 4=\frac12 \cdot 4 \cdot x \cdot \sin\,60^{\circ}+ \frac12 \cdot 3 \cdot x \cdot \sin\,30^{\circ} \end{gather} $$ (3) From there he calculated $x$: $$ \begin{gather} 6=2x \cdot \frac{\sqrt3}{2}+ \frac32 x \frac12 ~~~ /\cdot 4 \cr 24=4\sqrt3 x+3x \cr 24=x(4\sqrt3+3) ~~~/\cdot(4\sqrt3-3) \cr 24(4\sqrt3-3)=39x \cr x=\frac{32\sqrt3-24}{13} \end{gather} $$
Piotr demonstrated this solution:
(1) He drew a picture of a right-angled triangle.
(2) He used the Pythagorean theorem to find the length of hypotenuse $AB$: $$ |AB|=\sqrt{3^2+4^2}=5 $$
(3) Then he expressed: $$ \sin \alpha =\frac35,~\cos \alpha =\frac45 $$ (4) Finally, he used the Law of Sines to determine $x$: $$ \begin{gather} \frac{x}{\sin \alpha}=\frac{4}{\sin(120^{\circ}-\alpha)} \cr \frac53 x=\frac{4}{\sin 120^{\circ}\cos \alpha -\cos 120^{\circ}\sin \alpha} \cr \frac53 x=\frac{4}{\frac{\sqrt3}{2} \cdot \frac45+ \frac12 \cdot \frac35} / \cdot 3 \cr 5x=\frac{120}{4\sqrt3+3} \cr x=\frac{24}{4\sqrt3+3} \end{gather} $$ Their classmates commented on their solution. Which comment is right?
Both Tomek and Piotr demonstrated a correct solution.
Piotr's solution is correct. Tomek made a mistake in step (2). It should have been: $$ 6=2x \cdot \frac12+\frac32 x \frac{\sqrt3}{2} $$
Tomek's solution is correct. Piotr made a mistake in step (4). It should have been: $$ \frac53 x=\frac{4}{\frac{\sqrt3}{2} \cdot \frac45-\frac12 \cdot \frac35} $$
None of them showed a correct solution.
$$ \frac{24}{4\sqrt3+3} \cdot \frac{4 \sqrt3-3}{4\sqrt3-3}=\frac{32\sqrt3-24}{13} $$
