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

Answer

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Answer

$∠ K \approx {60.94}^{\circ}, ∠ C \approx {42.83}^{\circ}, ∠ A \approx {76.23}^{\circ}$

Steps

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 ： $\cos ∠ K = \frac{C^{2} + A^{2} - K^{2}}{2 C A} = \frac{7^{2} + 10^{2} - 9^{2}}{2 \times 7 \times 10}$

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

Step 5 ： $\cos ∠ C = \frac{K^{2} + A^{2} - C^{2}}{2 K A} = \frac{9^{2} + 10^{2} - 7^{2}}{2 \times 9 \times 10}$

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

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

Step 8 ： $∠ A = 180^{\circ} - ∠ K - ∠ C \approx 180^{\circ} - {60.94}^{\circ} - {42.83}^{\circ} \approx {76.23}^{\circ}$

Step 9 ： $∠ K \approx {60.94}^{\circ}, ∠ C \approx {42.83}^{\circ}, ∠ A \approx {76.23}^{\circ}$