\[ \begin{array}{|c|c|c|} \hline K & C & A \\ \hline \overline{9} & \overline{7} & \overline{10} \\ \hline \end{array} \] \begin{tabular}{|c|c|c|} \hline$K$ & $C$ & $A$ \\ \hline$\overline{9}$ & $\overline{7}$ & $\overline{10}$ \\ \hline \end{tabular} 07 cullah \[ \begin{array}{ll} =\frac{o p p}{h y p} \cos \theta=\frac{a d j}{h y p} \tan \theta=\frac{o p p}{a d j} & \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c} \\ 2+b^{2}-2 a b \cos C & \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \end{array} \] [2K] owledge he two triangles in the image below are similar. Determine the value of $x$ [2K] etermine the value of $\theta$, to the nearest degree. \[ \begin{array}{l} \theta=33.36 \\ \theta=33^{\circ} \end{array} \] Therefore, the value of $\varnothing$ 15 oppesise 7 is is $33^{\circ}$.

Step 1 ：Given a triangle with sides K = 9, C = 7, and A = 10, we need to find the angles of the triangle.

Step 2 ：Using the Law of Cosines, we can find the angles:

Step 3 ：\(\cos{\angle K} = \frac{C^2 + A^2 - K^2}{2CA} = \frac{7^2 + 10^2 - 9^2}{2 \times 7 \times 10}\)

Step 4 ：\(\angle K = \arccos{\left(\frac{7^2 + 10^2 - 9^2}{2 \times 7 \times 10}\right)} \approx 60.94^\circ\)

Step 5 ：\(\cos{\angle C} = \frac{K^2 + A^2 - C^2}{2KA} = \frac{9^2 + 10^2 - 7^2}{2 \times 9 \times 10}\)

Step 6 ：\(\angle C = \arccos{\left(\frac{9^2 + 10^2 - 7^2}{2 \times 9 \times 10}\right)} \approx 42.83^\circ\)

Step 7 ：Since the sum of angles in a triangle is 180°, we can find the third angle:

Step 8 ：\(\angle A = 180^\circ - \angle K - \angle C \approx 180^\circ - 60.94^\circ - 42.83^\circ \approx 76.23^\circ\)

Step 9 ：\(\boxed{\angle K \approx 60.94^\circ, \angle C \approx 42.83^\circ, \angle A \approx 76.23^\circ}\)