Area of triangle whose base is 12cm and angles at the base is 30 and 45 degree is

To find the area of the triangle with a base of 12 cm and angles at the base measuring 30° and 45°, we'll use trigonometry. Since the triangle has angles of 30°, 45°, and \(105^\circ\) (because the sum of the angles in a triangle is always \(180^\circ\)), we'll apply the formula:

\[
\text{Area} = \frac{1}{2} \times a \times b \times \sin C
\]

Where:
- \(a\) and \(b\) are the two sides adjacent to the angle \(C\),
- \(C\) is the angle opposite the base.

However, since we don't directly have the lengths of the two sides, we'll have to find them using the given base and angles.

### Step 1: Apply the Law of Sines to find the sides
Using the Law of Sines:
\[
\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}
\]
Here:
- \(c = 12 \, \text{cm}\) (base),
- \(A = 30^\circ\),
- \(B = 45^\circ\),
- \(C = 105^\circ\).

Let's calculate \(a\) and \(b\) (the other two sides):

\[
\frac{a}{\sin 45^\circ} = \frac{12}{\sin 105^\circ} \quad \text{and} \quad \frac{b}{\sin 30^\circ} = \frac{12}{\sin 105^\circ}
\]

Since:
\[
\sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 30^\circ = \frac{1}{2}, \quad \sin 105^\circ = \sin(180^\circ - 75^\circ) = \sin 75^\circ
\]

We calculate \( \sin 75^\circ \) using:
\[
\sin 75^\circ = \sin(45^\circ + 30^\circ) = \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ
\]

Now:
\[
\sin 75^\circ = \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}
\]

### Step 2: Calculate the area using trigonometry
Instead of finding the sides \(a\) and \(b\) explicitly, a more straightforward way to find the area is to use the following formula for the area of a triangle when the base and two angles are known:

\[
\text{Area} = \frac{1}{2} \times c^2 \times \frac{\sin A \times \sin B}{\sin C}
\]

Substituting the values:

\[
\text{Area} = \frac{1}{2} \times 12^2 \times \frac{\sin 30^\circ \times \sin 45^\circ}{\sin 105^\circ}
\]

We already know:
\[
\sin 30^\circ = \frac{1}{2}, \quad \sin 45^\circ = \frac{\sqrt{2}}{2}, \quad \sin 75^\circ = \frac{\sqrt{6} + \sqrt{2}}{4}
\]

So:
\[
\text{Area} = \frac{1}{2} \times 144 \times \frac{\frac{1}{2} \times \frac{\sqrt{2}}{2}}{\frac{\sqrt{6} + \sqrt{2}}{4}}
\]

Simplifying:
\[
\text{Area} = 72 \times \frac{\frac{\sqrt{2}}{4}}{\frac{\sqrt{6} + \sqrt{2}}{4}} = 72 \times \frac{\sqrt{2}}{\sqrt{6} + \sqrt{2}}
\]

To simplify further, multiply numerator and denominator by \( \sqrt{6} - \sqrt{2} \):

\[
\text{Area} = 72 \times \frac{\sqrt{2}(\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2})(\sqrt{6} - \sqrt{2})}
\]

This simplifies the denominator to \(4\), leading to:

\[
\text{Area} = 18 \times (\sqrt{12} - 2)
\]

Simplifying further:
\[
\text{Area} = 18 \times (2\sqrt{3} - 2) = 36\sqrt{3} - 36 \, \text{cm}^2
\]

So the area of the triangle is approximately \( \boxed{26.24 \, \text{cm}^2} \).

Would you like to go over any specific steps or explore related problems?

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Here are some related questions:

1. How does changing one of the base angles affect the area?
2. What is the area if both base angles are 45°?
3. How can you use the Law of Sines to solve for unknown side lengths in a triangle?
4. How does the Law of Cosines relate to this type of problem?
5. How do you find the height of the triangle from a different vertex using trigonometry?

**Tip:** Trigonometric identities and laws are powerful tools in solving problems where not all sides or angles are known, especially in non-right triangles.

Learn how to calculate the area of a triangle with a base of 12 cm and angles at the base measuring 30° and 45° using trigonometry and the Law of Sines. Explore step-by-step solutions and understand the application of trigonometric identities in determining triangle area. Suitable for high school students studying trigonometry.

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