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In the image you uploaded, two main questions are shown.

### Question 10:
You are given a right-angled triangle with sides:
- \(|AB| = 49.87\)
- \(|AC| = 58.6\)

You are asked to find angle \( \theta \) in radians to two decimal places.

To solve for \( \theta \), you can use the trigonometric function cosine, since \( \cos(\theta) = \frac{AB}{AC} \).

\[
\cos(\theta) = \frac{49.87}{58.6}
\]
Now solve for \( \theta \) using the inverse cosine:

\[
\theta = \cos^{-1}\left(\frac{49.87}{58.6}\right)
\]

### Question 11:
You are given the quadratic equation:
\[
6y^2 + 3.5y - 3 = 0
\]

The equation asks you to solve using the quadratic formula, which is:

\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 6 \), \( b = 3.5 \), and \( c = -3 \). You can now calculate the discriminant and substitute into the formula to find the solutions for \( y \).

I can work out these calculations for you. Would you like me to proceed?

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### 5 related questions:
1. How do you calculate angles in a triangle using sine, cosine, and tangent functions?
2. What is the quadratic formula and how is it derived?
3. How do radians relate to degrees, and how do you convert between them?
4. How do you interpret the discriminant of a quadratic equation?
5. Can you solve any triangle using only side lengths and trigonometric functions?

### Tip:
When solving for angles in a right triangle, always decide which trigonometric ratio to use (sine, cosine, or tangent) based on the sides given relative to the angle you’re solving for.

In the right-angled triangle shown, |AB|=49.87 and |AC|=58.6. Find the angle θ in radians (correct to two decimal places). Also, solve the quadratic equation 6y^2 + 3.5y - 3 = 0 using the quadratic formula.

This problem involves solving for the angle θ in a right-angled triangle using cosine and solving a quadratic equation using the quadratic formula. Learn step-by-step solutions for both trigonometric and algebraic equations. Ideal for students in Grades 9-11.

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