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image_path = '/mnt/data/1732388141655t2r0c36a.jpg' image = Image.open(image_path) image.show()successLet's solve these questions using the cosine rule.

Question 1:

To find the side $x$ in the triangle: The formula for the cosine rule is: $x^2 = a^2 + b^2 - 2ab\cos(C)$ Here:

• $a = 13 \, \text{cm}$,
• $b = 10 \, \text{cm}$,
• $C = 100^\circ$.

Substitute the values: $x^2 = 13^2 + 10^2 - 2(13)(10)\cos(100^\circ)$ $x^2 = 169 + 100 - 2(13)(10)(-0.1736) \, \text{(using } \cos(100^\circ) = -0.1736\text{)}.$ $x^2 = 169 + 100 + 45.136 = 314.136$ $x = \sqrt{314.136} \approx 17.7 \, \text{cm}.$

Answer: $x \approx 17.7 \, \text{cm}$ (to 1 decimal place).

Question 2:

To find side $BC$ in the triangle: Using the cosine rule: $BC^2 = AB^2 + AC^2 - 2(AB)(AC)\cos(\angle BAC)$ Here:

• $AB = 5 \, \text{m}$,
• $AC = 7 \, \text{m}$,
• $\angle BAC = 42^\circ$.

Substitute the values: $BC^2 = 5^2 + 7^2 - 2(5)(7)\cos(42^\circ)$ $BC^2 = 25 + 49 - 2(5)(7)(0.7431) \, \text{(using } \cos(42^\circ) = 0.7431\text{)}.$ $BC^2 = 25 + 49 - 52.017 = 21.983$ $BC = \sqrt{21.983} \approx 4.69 \, \text{m}.$

Answer: $BC \approx 4.69 \, \text{m}$ (to 3 significant figures).

Would you like further clarification or a detailed explanation of any steps?

Here are 5 related questions:

1. Can the cosine rule be applied to right-angled triangles? Why or why not?
2. How do you determine when to use the sine rule versus the cosine rule?
3. What is the significance of the angle being obtuse in the first question?
4. How does the value of the cosine change as the angle increases from $0^\circ$ to $180^\circ$?
5. Can the cosine rule be used to calculate angles as well? How?

Tip: Always double-check the cosine value (positive or negative) depending on whether the angle is acute or obtuse!