Area of a Triangle.

From Page 544 in your book, you have:
The area of a triangle equals one half the product of the lengths of any two sides and the sine of the angle between them.

This means that for an arbitrary triangle with an interior angle 𝜃, if sides of length a and b converge at an angle 𝜃, then you have the formula:
Area =
1
2
· a · b · sin(𝜃)

Use the formula above to answer the following. Remember that the longest side is opposite the largest angle.
Give exact answers. Decimal approximations will be marked wrong. Don't forget the degree symbol!
(a) A triangle has side lengths 6 cm and 15 cm. If the angle between these two sides is 53°, determine the area of the triangle.
Area =
cm2

(b) An obtuse triangle has an interior angle 109°. If the two shortest sides have lengths 7 cm and 19 cm, determine the area of the triangle.

Area =
cm2

(c) An obtuse triangle has an interior angle 125° and area 120cm2. If the shortest sides have lengths 7 cm and b cm, determine b in cm.

b =
cm

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!

Explore how to calculate the area of triangles using the formula Area = 1/2 * a * b * sin(θ). This page provides step-by-step solutions for triangles with given side lengths and angles, suitable for high school geometry and trigonometry students.

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