To solve the problems, we will use the formula for the area of a triangle:

$\text{Area} = \frac{1}{2} \cdot a \cdot b \cdot \sin(\theta)$

(a) Triangle with sides 6 cm and 15 cm, angle 53°

Given:

• $a = 6$ cm
• $b = 15$ cm
• $\theta = 53^\circ$

Calculating the area:

$\text{Area} = \frac{1}{2} \cdot 6 \cdot 15 \cdot \sin(53^\circ)$

Calculating $\sin(53^\circ)$:

$\sin(53^\circ) \approx 0.7986$

Now, substituting back into the area formula:

$\text{Area} = \frac{1}{2} \cdot 6 \cdot 15 \cdot 0.7986 = 45 \cdot 0.7986 \approx 35.927$

So, the area of the triangle is:

$\text{Area} \approx 35.927 \text{ cm}^2$

Exact answer (without approximation):

$\text{Area} = 45 \cdot \sin(53^\circ) \text{ cm}^2$

(b) Obtuse triangle with angle 109°, sides 7 cm and 19 cm

Given:

• $a = 7$ cm
• $b = 19$ cm
• $\theta = 109^\circ$

Calculating the area:

$\text{Area} = \frac{1}{2} \cdot 7 \cdot 19 \cdot \sin(109^\circ)$

Calculating $\sin(109^\circ)$:

$\sin(109^\circ) \approx 0.9455$

Substituting back into the area formula:

$\text{Area} = \frac{1}{2} \cdot 7 \cdot 19 \cdot 0.9455 \approx \frac{1}{2} \cdot 133 \cdot 0.9455 \approx 63.0065$

Exact answer (without approximation):

$\text{Area} = \frac{133}{2} \cdot \sin(109^\circ) \text{ cm}^2$

(c) Obtuse triangle with angle 125° and area 120 cm², sides 7 cm and $b$

Given:

• Area = 120 cm²
• $a = 7$ cm
• $\theta = 125^\circ$

Using the area formula:

$120 = \frac{1}{2} \cdot 7 \cdot b \cdot \sin(125^\circ)$

Calculating $\sin(125^\circ)$:

$\sin(125^\circ) \approx 0.8192$

Now substituting back into the equation:

$120 = \frac{1}{2} \cdot 7 \cdot b \cdot 0.8192$

Rearranging to find $b$:

$120 = \frac{7 \cdot b \cdot 0.8192}{2}$

Multiplying both sides by 2:

$240 = 7 \cdot b \cdot 0.8192$

Now solving for $b$:

$b = \frac{240}{7 \cdot 0.8192} \approx \frac{240}{5.7344} \approx 41.8$

Exact answer:

$b = \frac{240}{7 \sin(125^\circ)} \text{ cm}$

Summary of Answers:

(a) $\text{Area} = 45 \sin(53^\circ) \text{ cm}^2$
(b) $\text{Area} = \frac{133}{2} \sin(109^\circ) \text{ cm}^2$
(c) $b = \frac{240}{7 \sin(125^\circ)} \text{ cm}$

Would you like details on any specific part or have any questions? Here are some related questions to consider:

1. How would the area change if the angle in part (a) was increased?
2. What is the relationship between side lengths and angles in a triangle?
3. How can the sine rule help in finding unknown sides or angles?
4. Can you derive the area formula using Heron's formula?
5. What are some real-world applications of calculating the area of triangles?

Tip: Always double-check your sine values and use precise angles for more accurate results!