Math Problem Statement

help me solve this please

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.

Solution

Let's work through this problem step by step.

Part (a): Explain how you know that 1<x<91 < x < 9.

To determine the range of values for xx, 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=4cmAB = 4 \, \text{cm}
  • BC=(x+2)cmBC = (x + 2) \, \text{cm}
  • AC=7cmAC = 7 \, \text{cm}

Applying the triangle inequality theorem:

  1. AB+BC>ACAB + BC > AC
    4+(x+2)>74 + (x + 2) > 7
    x+6>7x + 6 > 7
    x>1x > 1

  2. AB+AC>BCAB + AC > BC
    4+7>(x+2)4 + 7 > (x + 2)
    11>x+211 > x + 2
    x<9x < 9

Thus, combining both inequalities, we have:

1<x<91 < x < 9

Part (b): Work out the value of xx and the area of the triangle for each case.

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

To find the value of xx, we can use the cosine rule:

BC2=AB2+AC22ABACcos(ABC)BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle ABC)

Substituting the given values:

(x+2)2=42+72247cos(60)(x + 2)^2 = 4^2 + 7^2 - 2 \cdot 4 \cdot 7 \cdot \cos(60^\circ)

(x+2)2=16+4956×0.5(x + 2)^2 = 16 + 49 - 56 \times 0.5

(x+2)2=6528(x + 2)^2 = 65 - 28

(x+2)2=37(x + 2)^2 = 37

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

x=3724.08x = \sqrt{37} - 2 \approx 4.08

Area of the triangle:

The area AA can be found using the formula:

A=12×AB×AC×sin(ABC)A = \frac{1}{2} \times AB \times AC \times \sin(\angle ABC)

A=12×4×7×sin(60)A = \frac{1}{2} \times 4 \times 7 \times \sin(60^\circ)

A=14×3214×0.86612.1cm2A = 14 \times \frac{\sqrt{3}}{2} \approx 14 \times 0.866 \approx 12.1 \, \text{cm}^2

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

Using the cosine rule again:

(x+2)2=42+72247cos(45)(x + 2)^2 = 4^2 + 7^2 - 2 \cdot 4 \cdot 7 \cdot \cos(45^\circ)

(x+2)2=16+4956×0.707(x + 2)^2 = 16 + 49 - 56 \times 0.707

(x+2)2=6539.6(x + 2)^2 = 65 - 39.6

(x+2)2=25.4(x + 2)^2 = 25.4

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

x=25.423.04x = \sqrt{25.4} - 2 \approx 3.04

Area of the triangle:

A=12×4×7×sin(45)A = \frac{1}{2} \times 4 \times 7 \times \sin(45^\circ)

A=14×2214×0.7079.90cm2A = 14 \times \frac{\sqrt{2}}{2} \approx 14 \times 0.707 \approx 9.90 \, \text{cm}^2

Summary:

  1. For ABC=60\angle ABC = 60^\circ, x4.08x \approx 4.08 and area 12.1cm2\approx 12.1 \, \text{cm}^2.
  2. For ABC=45\angle ABC = 45^\circ, x3.04x \approx 3.04 and area 9.90cm2\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.

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Math Problem Analysis

Mathematical Concepts

Triangle Inequality
Trigonometry
Cosine Rule
Sine Rule
Area of Triangles

Formulas

Cosine Rule: c^2 = a^2 + b^2 - 2ab * cos(C)
Sine Rule for Area: Area = 1/2 * a * b * sin(C)

Theorems

Triangle Inequality Theorem
Cosine Rule

Suitable Grade Level

Grades 10-12