https://file.aiquickdraw.com/mathbot/file/17259945165571ownuj6v.jpg

The image shows a right triangle \(ABC\) with:

- Angle \(\angle BAC = 35^\circ\).
- Side \(a = BC = 9\).
- Side \(AB = c\) (the hypotenuse).
- Side \(AC = b\) (the adjacent side to the \(35^\circ\) angle).

We need to find the lengths of sides \(b\) (adjacent to the \(35^\circ\) angle) and \(c\) (the hypotenuse).

### Step-by-Step Solution

1. **Find Side \(b\):**
   Using the tangent function, which is defined as:
   \[
   \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}
   \]
   For \(\theta = 35^\circ\):
   \[
   \tan(35^\circ) = \frac{a}{b} = \frac{9}{b}
   \]

   Solving for \(b\):
   \[
   b = \frac{9}{\tan(35^\circ)}
   \]

2. **Find Side \(c\):**
   Using the sine function:
   \[
   \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
   \]
   For \(\theta = 35^\circ\):
   \[
   \sin(35^\circ) = \frac{9}{c}
   \]

   Solving for \(c\):
   \[
   c = \frac{9}{\sin(35^\circ)}
   \]

Let's compute these values.The lengths of the sides of the triangle are:

- \(b \approx 12.85\) (the adjacent side to the \(35^\circ\) angle).
- \(c \approx 15.69\) (the hypotenuse).

Would you like further details or have any questions?

### Related Questions:
1. How do you calculate the area of this right triangle?
2. What would be the length of side \(b\) if the angle was \(45^\circ\) instead of \(35^\circ\)?
3. Can you use the cosine function to find the length of side \(b\)?
4. How do trigonometric functions relate to the unit circle?
5. What is the Pythagorean theorem, and how does it apply to this triangle?

**Tip:** Remember, trigonometric functions are crucial for solving right triangles when you know one side length and one angle.

Find the lengths of the sides b and c of a right triangle with a 35° angle, given side a = 9.

This guide explains how to calculate the lengths of the sides of a right triangle with a 35° angle, given side a = 9. Using trigonometric functions, we find side b (adjacent) and side c (hypotenuse) with step-by-step solutions. Suitable for students in Grades 9-11.

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