uppose a triangle has vertices A, B and C and has the following measurements: m∠C=91∘¯¯¯¯¯¯¯¯AC=40.1 cm¯¯¯¯¯¯¯¯BC=32.5 cm

Start by drawing a diagram of this triangle and labeling the known values.

What is the length of AB

in cm?

¯¯¯¯¯¯¯¯AB=

cm

What is the degree measure of ∠A

?

m∠A=

°

What is the degree measure of ∠B

?

m∠B=

°

Let's start by understanding the given problem:

1. **Vertices**: The triangle has vertices \(A\), \(B\), and \(C\).
2. **Angle at \(C\)**: \(m\angle C = 91^\circ\) (an obtuse angle).
3. **Side lengths**:
   - \(AC = 40.1\) cm
   - \(BC = 32.5\) cm
4. **Unknowns**:
   - The length of side \(AB\) (which we will call \(c\)).
   - The measures of angles \(A\) and \(B\).

### Step 1: Drawing the Diagram
We have a triangle with vertices \(A\), \(B\), and \(C\) where:
- \(AC = 40.1\) cm
- \(BC = 32.5\) cm
- \(m\angle C = 91^\circ\)

```
   A
   /\
  /  \
c/____\b
B       C
```

### Step 2: Applying the Law of Cosines
To find the length of side \(AB\) (denoted as \(c\)), we use the Law of Cosines:

\[
c^2 = a^2 + b^2 - 2ab \cdot \cos(C)
\]

Substituting the values:
\[
c^2 = (40.1)^2 + (32.5)^2 - 2 \times 40.1 \times 32.5 \times \cos(91^\circ)
\]

Note that \(\cos(91^\circ)\) is slightly negative because 91° is slightly greater than 90°.

### Step 3: Calculating \(c\)
Let's calculate \(c\):

\[
c^2 = 1608.01 + 1056.25 - 2 \times 40.1 \times 32.5 \times (-0.017452)
\]
\[
c^2 = 1608.01 + 1056.25 + 45.45
\]
\[
c^2 = 2709.71
\]
\[
c \approx \sqrt{2709.71} \approx 52.06 \text{ cm}
\]

So, \( \overline{AB} \approx 52.06 \text{ cm} \).

### Step 4: Finding Angles \(A\) and \(B\)
Now, let's use the Law of Sines to find the remaining angles.

The Law of Sines states:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]

1. **Find \(m\angle A\):**
\[
\sin A = \frac{a \cdot \sin C}{c}
\]
\[
\sin A = \frac{40.1 \times \sin(91^\circ)}{52.06} \approx \frac{40.1 \times 1}{52.06} \approx 0.7704
\]
\[
m\angle A \approx \arcsin(0.7704) \approx 50.3^\circ
\]

2. **Find \(m\angle B\):**
Since the sum of angles in a triangle is \(180^\circ\):
\[
m\angle B = 180^\circ - m\angle C - m\angle A
\]
\[
m\angle B = 180^\circ - 91^\circ - 50.3^\circ \approx 38.7^\circ
\]

### Final Results:
- \( \overline{AB} \approx 52.06 \) cm
- \( m\angle A \approx 50.3^\circ \)
- \( m\angle B \approx 38.7^\circ \)

---

Would you like a more detailed explanation on any of the steps? Here are some related questions to explore:

1. How is the Law of Cosines derived?
2. What is the significance of the Law of Sines in solving triangles?
3. How can you determine if a triangle is acute, obtuse, or right using side lengths?
4. Why does the sine of angles greater than \(90^\circ\) result in a negative value?
5. How do you calculate the area of a triangle using the given sides and angle?
6. Can the Law of Cosines be used to solve any type of triangle?
7. What would happen if one of the angles was given as exactly \(90^\circ\)?
8. How does changing one side of the triangle affect the angles and the other sides?

**Tip**: When dealing with angles slightly above \(90^\circ\), always check the sign of trigonometric functions, as it affects calculations significantly.

Explore how to find the side length and angles of a triangle with given measurements using the Law of Cosines and Law of Sines. Discover step-by-step solutions for finding side AB, angle A, and angle B.

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