Question

2.2 In the diagram, ∆𝐴𝐵𝐶 are given with 𝐴 = 39°, 𝐶 = 46°, 𝐵𝐶 = 2𝑐𝑚 and
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:
\[
\boxed{2.3}
\]

---

### 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:
\[
\boxed{8.2}
\]