Derive a formula for the area of the triangle, given b, A, B, and C. That is, derive the formula: (b^2sinAsinC)/(2sinB)

To derive the formula for the area of a triangle given the base \( b \) and angles \( A \), \( B \), and \( C \), we can utilize the sine rule and the definition of the area of a triangle. Let's break this down step by step.

Understanding Triangle Area

The area \( A \) of a triangle can generally be calculated using the formula:

Area = 0.5 × base × height

However, when we have angles and one side, we can use a different approach involving the sine of the angles.

Using the Sine Rule

The sine rule states that for any triangle with sides \( a \), \( b \), and \( c \) opposite to angles \( A \), \( B \), and \( C \) respectively, the following relationship holds:

\( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)

From this, we can express the sides in terms of the angles and the known side \( b \):

\( a = \frac{b \cdot \sin A}{\sin B} \)

\( c = \frac{b \cdot \sin C}{\sin B} \)

Finding the Area

Now, we can use the formula for the area of a triangle in terms of two sides and the included angle. The area can also be expressed as:

Area = 0.5 × a × c × \sin B

Substituting the expressions for \( a \) and \( c \) from the sine rule, we get:

Area = 0.5 × \left(\frac{b \cdot \sin A}{\sin B}\right) × \left(\frac{b \cdot \sin C}{\sin B}\right) × \sin B

Simplifying the Expression

Now, let's simplify this expression:

• First, notice that one \( \sin B \) in the denominator cancels with \( \sin B \) in the numerator:

Area = 0.5 × \frac{b^2 \cdot \sin A \cdot \sin C}{\sin B}

Multiplying both sides by 2 gives us:

Area = \frac{b^2 \cdot \sin A \cdot \sin C}{2 \cdot \sin B}

Final Formula

Thus, the derived formula for the area of the triangle, given the base \( b \) and angles \( A \), \( B \), and \( C \), is:

Area = \frac{b^2 \cdot \sin A \cdot \sin C}{2 \cdot \sin B}

Practical Example

For instance, if you have a triangle where \( b = 10 \), \( A = 30^\circ \), \( B = 60^\circ \), and \( C = 90^\circ \), you can substitute these values into the formula:

• Calculate \( \sin A = \sin 30^\circ = 0.5\)
• Calculate \( \sin C = \sin 90^\circ = 1\)
• Calculate \( \sin B = \sin 60^\circ \approx 0.866\)

Plugging these into the formula gives:

Area = \frac{10^2 \cdot 0.5 \cdot 1}{2 \cdot 0.866} \approx 28.87

This process illustrates how you can derive the area of a triangle using angles and one side, providing a powerful tool for solving various problems in geometry.