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We are given a right-angled triangle \( ABC \), where:

- \( AB = 3.5 \, \text{cm} \)
- \( BC = 2.3 \, \text{cm} \)
- \( \angle B = 90^\circ \)

We need to find \( \angle C \) correct to one decimal place.

Since the triangle is right-angled at \( B \), we can use trigonometric ratios to find \( \angle C \). The tangent of angle \( C \) is the ratio of the opposite side to the adjacent side.

\[
\tan(\angle C) = \frac{\text{opposite}}{\text{adjacent}} = \frac{BC}{AB}
\]

Substituting the values:

\[
\tan(\angle C) = \frac{2.3}{3.5}
\]

Now, let's calculate this and find the angle \( C \).The angle \( \angle C \) is approximately \( 33.3^\circ \), correct to one decimal place.

Let me know if you need further details or clarifications.

Here are 5 related questions to expand your understanding:
1. How would you find \( \angle A \) using the information given?
2. What is the sine of \( \angle C \) based on the triangle's dimensions?
3. How can you calculate the hypotenuse of the triangle using the Pythagorean theorem?
4. What other trigonometric ratio could you use to find \( \angle C \)?
5. How would you solve for \( \angle C \) if only the hypotenuse was given?

**Tip:** Always remember to check that your calculator is in the correct mode (degrees or radians) when working with angles!

A right-angled triangle ABC is such that AB = 3.5 cm, BC = 2.3 cm, and angle B = 90°. Find angle C correct to one decimal place.

Learn how to find angle C in a right-angled triangle where AB = 3.5 cm, BC = 2.3 cm, and angle B = 90°. This problem involves applying the tangent ratio to calculate the angle to one decimal place. Suitable for students in Grades 9-11.

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