Let's solve these triangle problems one by one:

1. Calculate the area of the triangle $ABC$:

• Given: $AB = 7 \, \text{cm}$, $AC = 12 \, \text{cm}$, and angle $\angle ACB = 77^\circ$.
• The formula for the area of a triangle using two sides and the included angle is:

$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle ACB)$

$\text{Area} = \frac{1}{2} \times 7 \times 12 \times \sin(77^\circ) = 42 \times 0.9744 = 40.93 \, \text{cm}^2$

So, the area is approximately 40.93 cm².

2. Find the area of isosceles triangle $XYZ$:

• Given: $XY = YZ = 9.7 \, \text{km}$, $XZ = 11.8 \, \text{km}$, and angle $\angle XYZ = 45^\circ$.
• The formula for the area is:

$\text{Area} = \frac{1}{2} \times XY \times XZ \times \sin(\angle XYZ)$

$\text{Area} = \frac{1}{2} \times 9.7 \times 11.8 \times \sin(45^\circ) = 57.23 \times 0.7071 = 40.46 \, \text{km}^2$

So, the area is approximately 40.46 km².

3. Calculate the area of triangle $ABC$:

• Given: $AB = 19.1 \, \text{cm}$, $AC = 15.8 \, \text{cm}$, and angle $\angle ACB = 22^\circ$.
• Using the area formula for two sides and the included angle:

$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle ACB)$

$\text{Area} = \frac{1}{2} \times 19.1 \times 15.8 \times \sin(22^\circ) = 150.19 \times 0.3746 = 56.24 \, \text{cm}^2$

So, the area is approximately 56.24 cm².

4. Find the length $x$ given the area of the triangle is $30 \, \text{cm}^2$:

• Given: $\text{Area} = 30 \, \text{cm}^2$, $\angle = 38^\circ$, and base $= 12 \, \text{cm}$.
• Using the area formula:

$\text{Area} = \frac{1}{2} \times 12 \times x \times \sin(38^\circ)$

$30 = 6 \times x \times 0.6157 \quad \Rightarrow \quad x = \frac{30}{6 \times 0.6157} = \frac{30}{3.694} \approx 8.12 \, \text{cm}$

So, the length $x$ is approximately 8.12 cm.

5. Find the angle $x$ given the area is $42 \, \text{cm}^2$:

• Given: $\text{Area} = 42 \, \text{cm}^2$, $AB = 14 \, \text{cm}$, $BC = 19 \, \text{cm}$.
• Using the area formula:

$\text{Area} = \frac{1}{2} \times 14 \times 19 \times \sin(x)$

$42 = 133 \times \sin(x) \quad \Rightarrow \quad \sin(x) = \frac{42}{133} = 0.315$

Now, find the angle:

$x = \arcsin(0.315) \approx 18.36^\circ$

So, the angle $x$ is approximately 18.36°.

Would you like more details on any of the steps?

Here are 5 additional questions to expand your understanding:

1. How can the sine rule be applied in triangle area calculations?
2. What other methods can be used to find the area of triangles besides sine?
3. What are the key differences between solving for angles and solving for sides in triangles?
4. How does the unit (cm, km, etc.) affect the final answers in area calculations?
5. How would the calculations change if the triangles were right-angled?

Tip:

Remember to always check if a triangle is right-angled, as it simplifies the calculations significantly!