2.2 In the diagram, ∆𝐴𝐵𝐶 are given with 𝐴 = 39°, 𝐶 = 46°, 𝐵𝐶 = 2𝑐𝑚 and

Question

𝐴𝐵 = 𝑥𝑐𝑚.
Figure 2: Diagram for Question 2.2.
Determine 𝑥, round of your answer to one decimal place. (4)
2.3 In the diagram, ∆𝐴𝐵𝐶 are given with 𝐶 = 60°, 𝐴𝐶 = 9𝑐𝑚 and 𝐵𝐶 = 7𝑐𝑚.
Figure 3: Diagram for Question 2.3.
Determine 𝐴𝐵, round of your answer to one decimal place.

Answer 1

To solve the questions, we will use the Law of Sines for both triangles.

For Question 2.2:

Given:

• Angle A = 39°
• Angle C = 46°
• Side BC = 2 cm
• Side AB = x cm

First, we find angle B: \[ \text{Angle B} = 180° - \text{Angle A} - \text{Angle C} = 180° - 39° - 46° = 95° \]

Now we apply the Law of Sines: \[ \frac{AB}{\sin C} = \frac{BC}{\sin A} \] This can be set up as: \[ \frac{x}{\sin(46°)} = \frac{2}{\sin(39°)} \]

Now, we can solve for x: \[ x = \frac{2 \cdot \sin(46°)}{\sin(39°)} \]

Calculating the values: \[ \sin(46°) \approx 0.7193 \] \[ \sin(39°) \approx 0.6293 \]

Now substituting these values into the equation: \[ x = \frac{2 \cdot 0.7193}{0.6293} \approx \frac{1.4386}{0.6293} \approx 2.28 \]

Rounding to one decimal place: \[ \boxed{2.3} \]

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For Question 2.3:

Given:

• Angle C = 60°
• Side AC = 9 cm
• Side BC = 7 cm

First, we need to find angle A. We will again apply the Law of Sines: \[ \frac{AC}{\sin B} = \frac{BC}{\sin C} \] We need to find angle B first: Since the angles in a triangle add up to 180°: \[ \text{Angle A} + \text{Angle B} + 60° = 180° \] \[ \text{Angle A} + \text{Angle B} = 120° \] This means: \[ \text{Angle B} = 120° - \text{Angle A} \]

Now applying the Law of Sines: \[ \frac{9}{\sin(120° - A)} = \frac{7}{\sin(60°)} \]

Since \(\sin(60°) = \frac{\sqrt{3}}{2} \approx 0.8660\), we can substitute: \[ \frac{9}{\sin(120° - A)} = \frac{7}{0.8660} \]

Cross-multiply: \[ 9 \cdot 0.8660 = 7 \cdot \sin(120° - A) \] \[ 7 \cdot \sin(120° - A) \approx 7.794 \] Now solving for \(\sin(120° - A)\): \[ \sin(120° - A) \approx \frac{7.794}{7} \approx 1.114 \]

Since the sin value cannot exceed 1, we need to consider the other approach: We continue from the Law of Sines directly: Taking angle A directly. We can find angle B using side ratios since B + A = 120°.

Using trial and error or numerical methods, the Law of Cosines might be a better approach:

Using the known sides and the angle:

1. Calculate using Law of Cosines: For side AB: \[ AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(60°) \] Substituting values gives: \[ AB^2 = 9^2 + 7^2 - 2 \cdot 9 \cdot 7 \cdot 0.5 \] \[ = 81 + 49 - 63 \] \[ = 67 \] Thus, \[ AB = \sqrt{67} \approx 8.2 \] Therefore, round off: \[ \boxed{8.2} \]