91Ó°ÊÓ

The area of a triangle is a key measurement that helps determine how much space is contained within the triangle's boundaries. One common way to calculate this is using the triangle area formula:

• \( A = \frac{1}{2}ab\sin(C) \)

Here, \( a \) and \( b \) are the lengths of two sides, and \( C \) is the angle between these sides. This formula is particularly useful for non-right triangles, such as the isosceles triangle in our exercise.
Applying this formula requires knowledge of trigonometry since it involves the sine function, which helps relate the angle's value to the triangle's sides and area. In our problem, the isosceles triangle uses this formula to marry the lengths of the equal sides (which we denote as \( x \)) to the given area of \( 24 \text{ cm}^2 \) and the angle \( C = \frac{5\pi}{6} \). This approach allows for a seamless calculation of geometric properties even when not all traditional measures like height are instantly available.

Trigonometry is integral in solving triangle-related problems involving angles and lengths. It provides the tools for connecting angles with sides through functions like sine, cosine, and tangent. In this exercise, the sine function is particularly significant.
We use \( \sin\left(\frac{5\pi}{6}\right) \), where \( \frac{5\pi}{6} \) is an angle in radians, equivalent to \( 150^\circ \).
To understand this, the sine of \( 150^\circ \) is \( \sin\left(150^\circ\right) = \frac{1}{2} \), which is calculated based on the unit circle or from trigonometric tables.
This value is important because it tells us how the angle interacts with the side lengths of the triangle when trying to find the area. By knowing \( \sin\left(\frac{5\pi}{6}\right) \), we effectively convert the angle into a usable multiplier in our area formula. Understanding these fundamental trigonometric values is key when dealing with non-right-angle triangles in geometry.

Solving equations is the process of finding unknown values that satisfy a given mathematical statement. In our triangle problem, after applying the area formula and substituting known values, we encounter an equation to solve for \( x \), where \( x \) is the length of the equal sides of the isosceles triangle.
Let's recap the steps:

• Start with: \( 24 = \frac{1}{2} x^2 \sin\left(\frac{5\pi}{6}\right) \).
• Substitute \( \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \), leading to \( 24 = \frac{1}{4} x^2 \).
• Clear the fraction by multiplying both sides by 4, resulting in \( x^2 = 96 \).
• Finally, solve for \( x \) by taking the square root of both sides: \( x = \sqrt{96} = 4\sqrt{6} \).

Breaking down the equation like this ensures clarity and reveals how each mathematical operation contributes to finding the solution. By isolating \( x^2 \) and solving step by step, we unravel the original problem into manageable parts. This method is crucial for methodically determining unknowns in math problems.