Clearly $AC = 5.7$. Now coordinate bash?

bruh no coordinate bash law of cosines then find area with sine area - divide by bases to get heights (dont have time to actually do it rn)

parmenides51 wrote: The triangle $ABC$ has $\angle ACB = 30^o$, $BC = 4$ cm and $AB = 3$ cm . Compute the altitudes of the triangle. Translation from the official solution by ChatGPT: Given a circle \( \mathcal{C} \) with center \( B \) and radius 3. In the half-plane bounded by the line segment \( BC \) that contains point \( A \), construct the ray \( CT \) such that \(\angle BCT = 30^\circ\). If \( M \) is the projection of point \( B \) onto line \( CT \), then from the right triangle \( BMC \) we deduce \( BM = 2 \), therefore \( CT \) is a secant of the circle \( \mathcal{C} \). If \(\{A_1, A_2\} = \mathcal{C} \cap CT\), then \( A \in \{A_1, A_2\} \), $CA_1<CA_2$. Case 1: \( A = A_1 \). Since \( MA_1 = \sqrt{5} \) and \( CM = 2\sqrt{3} \), it results \( A_1 C = 2\sqrt{3} - \sqrt{5} \). If \( A_1 N_1 \perp BC \), \( N_1 \in BC \), then since \(\angle BCA = 30^\circ \), it results \( A_1 N_1 = \frac{A_1 C}{2} = \frac{2\sqrt{3} - \sqrt{5}}{2} \). If \( C C_1 \perp BA_1 \), \( C_1 \in BA_1 \), then $C C_1 \cdot BA_1 = BM \cdot A_1 C \implies C C_1 = \frac{2 (2\sqrt{3} - \sqrt{5})}{3}$ Case 2: \( A = A_2 \). Since \( CA_2 = CM + MA_2 \), it results \( CA_2 = 2\sqrt{3} + \sqrt{5} \). Proceeding similarly as in Case 1, we obtain that the length of the altitude from \( A_2 \) is $\frac{2\sqrt{3} + \sqrt{5}}{2}$ and the corresponding value for \( C \) is $\frac{2(2\sqrt{3} + \sqrt{5})}{3}$ In both cases, the length of the altitude from \( B \) is 2.

what the heck wrote: Given a circle \( \mathcal{C} \) with center \( B \) and radius 3. In the half-plane bounded by the line segment \( BC \) that contains point \( A \), construct the ray \( CT \) such that \(\angle BCT = 30^\circ\). If \( M \) is the projection of point \( B \) onto line \( CT \), then from the right triangle \( BMC \) we deduce \( BM = 2 \), therefore \( CT \) is a secant of the circle \( \mathcal{C} \). If \(\{A_1, A_2\} = \mathcal{C} \cap CT\), then \( A \in \{A_1, A_2\} \), $CA_1<CA_2$. Case 1: \( A = A_1 \). Since \( MA_1 = \sqrt{5} \) and \( CM = 2\sqrt{3} \), it results \( A_1 C = 2\sqrt{3} - \sqrt{5} \). If \( A_1 N_1 \perp BC \), \( N_1 \in BC \), then since \(\angle BCA = 30^\circ \), it results \( A_1 N_1 = \frac{A_1 C}{2} = \frac{2\sqrt{3} - \sqrt{5}}{2} \). If \( C C_1 \perp BA_1 \), \( C_1 \in BA_1 \), then $C C_1 \cdot BA_1 = BM \cdot A_1 C \implies C C_1 = \frac{2 (2\sqrt{3} - \sqrt{5})}{3}$ Case 2: \( A = A_2 \). Since \( CA_2 = CM + MA_2 \), it results \( CA_2 = 2\sqrt{3} + \sqrt{5} \). Proceeding similarly as in Case 1, we obtain that the length of the altitude from \( A_2 \) is $\frac{2\sqrt{3} + \sqrt{5}}{2}$ and the corresponding value for \( C \) is $\frac{2(2\sqrt{3} + \sqrt{5})}{3}$ In both cases, the length of the altitude from \( B \) is 2. what the overkill