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Let's work through this problem step by step.

### **Part (a): Explain how you know that \( 1 < x < 9 \).**

To determine the range of values for \( x \), we need to use the triangle inequality theorem. This theorem states that, in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.

Given:

- \( AB = 4 \, \text{cm} \)
- \( BC = (x + 2) \, \text{cm} \)
- \( AC = 7 \, \text{cm} \)

Applying the triangle inequality theorem:

1. \( AB + BC > AC \)  
   \( 4 + (x + 2) > 7 \)  
   \( x + 6 > 7 \)  
   \( x > 1 \)

2. \( AB + AC > BC \)  
   \( 4 + 7 > (x + 2) \)  
   \( 11 > x + 2 \)  
   \( x < 9 \)

Thus, combining both inequalities, we have:

\[
1 < x < 9
\]

### **Part (b): Work out the value of \( x \) and the area of the triangle for each case.**

#### **Case (i): \( \angle ABC = 60^\circ \)**

To find the value of \( x \), we can use the cosine rule:

\[
BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle ABC)
\]

Substituting the given values:

\[
(x + 2)^2 = 4^2 + 7^2 - 2 \cdot 4 \cdot 7 \cdot \cos(60^\circ)
\]

\[
(x + 2)^2 = 16 + 49 - 56 \times 0.5
\]

\[
(x + 2)^2 = 65 - 28
\]

\[
(x + 2)^2 = 37
\]

\[
x + 2 = \sqrt{37}
\]

\[
x = \sqrt{37} - 2 \approx 4.08
\]

**Area of the triangle:**

The area \( A \) can be found using the formula:

\[
A = \frac{1}{2} \times AB \times AC \times \sin(\angle ABC)
\]

\[
A = \frac{1}{2} \times 4 \times 7 \times \sin(60^\circ)
\]

\[
A = 14 \times \frac{\sqrt{3}}{2} \approx 14 \times 0.866 \approx 12.1 \, \text{cm}^2
\]

#### **Case (ii): \( \angle ABC = 45^\circ \)**

Using the cosine rule again:

\[
(x + 2)^2 = 4^2 + 7^2 - 2 \cdot 4 \cdot 7 \cdot \cos(45^\circ)
\]

\[
(x + 2)^2 = 16 + 49 - 56 \times 0.707
\]

\[
(x + 2)^2 = 65 - 39.6
\]

\[
(x + 2)^2 = 25.4
\]

\[
x + 2 = \sqrt{25.4}
\]

\[
x = \sqrt{25.4} - 2 \approx 3.04
\]

**Area of the triangle:**

\[
A = \frac{1}{2} \times 4 \times 7 \times \sin(45^\circ)
\]

\[
A = 14 \times \frac{\sqrt{2}}{2} \approx 14 \times 0.707 \approx 9.90 \, \text{cm}^2
\]

### Summary:

1. For \( \angle ABC = 60^\circ \), \( x \approx 4.08 \) and area \( \approx 12.1 \, \text{cm}^2 \).
2. For \( \angle ABC = 45^\circ \), \( x \approx 3.04 \) and area \( \approx 9.90 \, \text{cm}^2 \).

Would you like more details, or have any questions?

### Related Questions:

1. How does the cosine rule apply in non-right triangles?
2. What is the significance of the triangle inequality theorem?
3. How do different angles affect the side lengths in a triangle?
4. Can you derive the sine and cosine values for common angles?
5. How can the area of a triangle be determined using different formulas?

### Tip: 

Always double-check angle measures and side calculations when using trigonometric functions to ensure accuracy in your results.

In ∆ABC, AB = 4 cm, BC = (x + 2) cm, and AC = 7 cm.
a) Explain how you know that 1 < x < 9.
b) Work out the value of x and the area of the triangle for the cases when i) ∠ABC = 60° and ii) ∠ABC = 45°, giving your answers to 3 significant figures.

This problem explores triangle inequalities and trigonometric concepts. Learn how to find x and calculate the area for different angles using the cosine rule and sine rule for a triangle with sides AB = 4 cm, BC = (x + 2) cm, and AC = 7 cm. Solutions provided for ∠ABC = 60° and ∠ABC = 45°, with detailed steps. Suitable for students in Grades 10-12.

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