We can solve for angle $\theta$ in the triangle using the Cosine Rule. The cosine rule states:

$\cos(\theta) = \frac{a^2 + b^2 - c^2}{2ab}$

where:

• $a$, $b$, and $c$ are the sides of the triangle,
• $\theta$ is the angle opposite side $c$.

Step 1: Label the triangle

From the image:

• Side $AC = 75 \, \text{cm}$ is opposite angle $\theta$,
• Side $AB = 52 \, \text{cm}$,
• Side $BC = 41 \, \text{cm}$.

Step 2: Apply the cosine rule

Here, $a = 52$, $b = 41$, and $c = 75$.

Using the cosine rule: $\cos(\theta) = \frac{52^2 + 41^2 - 75^2}{2 \times 52 \times 41}$

Step 3: Calculate the values

$52^2 = 2704, \quad 41^2 = 1681, \quad 75^2 = 5625$

$\cos(\theta) = \frac{2704 + 1681 - 5625}{2 \times 52 \times 41}$ $\cos(\theta) = \frac{4305 - 5625}{4264}$ $\cos(\theta) = \frac{-1320}{4264}$ $\cos(\theta) = -0.3096$

Step 4: Find $\theta$

Now, use the inverse cosine function to find $\theta$:

$\theta = \cos^{-1}(-0.3096)$

$\theta \approx 108^\circ$

Final Answer:

$\theta \approx 108^\circ$

Let me know if you need more details or further questions! Here are five relative questions that might help:

1. What other ways can we calculate the angle in a triangle if all sides are known?
2. How does the cosine rule differ from the sine rule?
3. Could this triangle's angles also be solved using trigonometry ratios?
4. What would the angle sum of this triangle be once we find all angles?
5. How do we find the remaining two angles in this triangle?

Tip: The cosine rule is especially useful in non-right triangles!