To derive the area of triangle $\triangle ABC$ using trigonometric ratios, let's start with the information given in the diagram.

Given:

• $\theta$ is the angle at $C$.
• $a = BC$ (opposite side of $\theta$).
• $b = CA$ (adjacent side to $\theta$).
• $h =$ height from $B$ perpendicular to $CA$.

Goal:

Derive the formula for the area of $\triangle ABC$ in terms of $a$, $b$, and $\theta$.

Step 1: Express the area using base and height.

The area $A$ of the triangle is given by: $A = \frac{1}{2} \times \text{base} \times \text{height}$ Here, the base is $b$, and the height is $h$. So, $A = \frac{1}{2} \times b \times h$

Step 2: Express the height $h$ using trigonometry.

In the right triangle $\triangle BCA$, the height $h$ can be expressed using the sine function: $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{a}$ Thus, $h = a \sin(\theta)$

Step 3: Substitute $h$ into the area formula.

Now, substitute $h = a \sin(\theta)$ into the area formula: $A = \frac{1}{2} \times b \times a \sin(\theta)$

Final Formula:

$\boxed{A = \frac{1}{2} ab \sin(\theta)}$ This is the formula for the area of a triangle using the lengths of two sides $a$ and $b$ and the sine of the included angle $\theta$.

Would you like further details, or do you have any questions?

Here are some related questions you might find interesting:

1. How can you derive the area formula for a triangle using other trigonometric functions like cosine or tangent?
2. How does the area formula change if $\theta$ is a right angle?
3. What happens to the area formula if one of the sides of the triangle is the hypotenuse of a right triangle?
4. Can the formula $A = \frac{1}{2} ab \sin(\theta)$ be used for non-acute triangles?
5. How can you derive the area of a triangle using Heron's formula, and how does it compare to this method?

Tip: When dealing with trigonometric ratios, always ensure the angle and sides correspond correctly in the right triangle. This avoids errors in calculations or derivations.