The two non-parallel sides of an isosceles trapezoid are each 7 feet long. The longer of the two bases measures 22 feet long. The sum of the base angles is 140°.

A. Use the Law of Cosines to find the length of the diagonal.
B. Use the Law of Sines to find the length of the shorter base.
Round your answers to the nearest hundredth.

To solve this problem, we'll follow the given steps. Here's the plan:

1. **Using the Law of Cosines to find the length of the diagonal.**
2. **Using the Law of Sines to find the length of the shorter base.**

Let's start with part A.

### Part A: Using the Law of Cosines to Find the Length of the Diagonal

Given:
- The non-parallel sides (legs) \( AD \) and \( BC \) are each 7 feet.
- The longer base \( AB \) is 22 feet.
- The sum of the base angles \( \angle DAB \) and \( \angle ABC \) is 140°.

Since the sum of the base angles is 140°, each base angle is \( \frac{140°}{2} = 70° \).

Let's denote the diagonal as \( AC \).

In \( \triangle ACD \), we can use the Law of Cosines. To apply this, we need to know the angle \( \angle CAD \).

\[ \angle CAD = 180° - 70° = 110° \]

Using the Law of Cosines:

\[ AC^2 = AD^2 + DC^2 - 2 \cdot AD \cdot DC \cdot \cos(\angle CAD) \]

Given:
- \( AD = 7 \) feet
- \( DC = 22 - \text{shorter base} \) feet
- \( \angle CAD = 110° \)

First, we need to approximate the length of \( DC \), which we'll denote as \( b \).

Substituting the known values:

\[ AC^2 = 7^2 + 7^2 - 2 \cdot 7 \cdot 7 \cdot \cos(110°) \]

\[ AC^2 = 49 + 49 - 2 \cdot 49 \cdot \cos(110°) \]

\[ AC^2 = 98 - 98 \cdot (-0.342) \]

\[ AC^2 = 98 + 33.516 \]

\[ AC^2 = 131.516 \]

\[ AC = \sqrt{131.516} \]

\[ AC \approx 11.47 \]

### Part B: Using the Law of Sines to Find the Length of the Shorter Base

To find the shorter base, we will denote it as \( b \). 

The diagonal divides the trapezoid into two triangles, \( \triangle ABD \) and \( \triangle BCD \).

Using the Law of Sines:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]

Here \( a \) and \( b \) are the sides, \( c \) is the base, and \( A \), \( B \), and \( C \) are the angles. We already know that:

\[ AC = 11.47 \]

Using \( \triangle ACD \) and knowing \( \angle CAD = 110° \):

\[ \frac{AD}{\sin(\angle ACD)} = \frac{CD}{\sin(\angle DAC)} \]

Given:
- \( AD = 7 \)
- \( \angle ACD = 70° \)
- \( \angle CAD = 110° \)

Using the Law of Sines:

\[ \frac{7}{\sin(70°)} = \frac{b}{\sin(110°)} \]

We know:

\[ \sin(70°) \approx 0.9397 \]
\[ \sin(110°) \approx 0.9397 \]

Since \( \sin(70°) = \sin(110°) \):

\[ \frac{7}{0.9397} = \frac{b}{0.9397} \]

This means:

\[ b = 7 \]

Therefore, the length of the shorter base is 7 feet.

So, the length of the diagonal \( AC \) is approximately \( 11.47 \) feet, and the length of the shorter base is \( 7 \) feet.

Would you like more details or have any questions?

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Here are some related questions to expand your understanding:

1. What is the formula for the Law of Cosines and when is it used?
2. How does the Law of Sines differ from the Law of Cosines?
3. What are the properties of an isosceles trapezoid?
4. How do you find the area of an isosceles trapezoid?
5. Can you derive the diagonal length for any trapezoid using similar methods?
6. What are the different types of trapezoids, and how do their properties differ?
7. How can trigonometric identities be used to simplify calculations involving angles?
8. How do the angles of a trapezoid affect its side lengths and diagonals?

**Tip:** Always verify the given conditions and known values before applying trigonometric formulas to avoid calculation errors.

Explore how to solve geometry problems involving isosceles trapezoids using the Law of Cosines and Law of Sines. This detailed solution covers finding diagonal length and base lengths with step-by-step calculations rounded to the nearest hundredth. Suitable for high school students learning geometry and trigonometry.

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