Problem 124 - Algebra and Trigonometry Real Mathematics, Real People

Question

The length of each of the two equal sides of an isosceles triangle is 10 meters (see figure). The angle between the two sides is \(\theta\) (a) Write the area of the triangle as a function of \(\theta / 2\). (b) Write the area of the triangle as a function of \(\theta\) and determine the value of \(\theta\) such that the area is a maximum.

Step-by-Step Solution

Answer

The area of the triangle can be written as \(Area = 100\sin(\theta / 2)\cos(\theta / 2)\) square meters when expressed as a function of \(\theta / 2\) and as \(Area = 50\sin(\theta)\) square meters as a function of \(\theta\). The area is at a maximum when \(\theta = \frac{\pi}{2}\) radians or 90 degrees.

Step 1: Write the formula for the area

The area of a triangle when two sides and the included angle are given is calculated by the formula \( Area = 0.5ab\sin(c) \) where \(a\) and \(b\) are the lengths of the two sides and \(c\) is the included angle. For our triangle, both \(a\) and \(b\) are 10 meters, and \(c\) is \(\theta\). Therefore the area \(A\) will be a function of \(\theta\), so we have \( Area = 0.5(10)(10)\sin(\theta) = 50\sin(\theta) \) sq. meters.

Step 2: Write the area as a function of \(\theta / 2\)

By the double angle formula, we know that \(\sin(\theta) = 2\sin(\theta / 2)\cos(\theta / 2)\). Hence, there will be the new equation for the area as \( Area = 50 * 2\sin(\theta / 2)\cos(\theta / 2) = 100\sin(\theta / 2)\cos(\theta / 2) \) sq. meters.

Step 3: Find the maximum area as a function of \(\theta\)

The function \(Area = 50\sin(\theta)\) reaches its maximum at \(\theta = \frac{\pi}{2}\) radians, which is also 90 degrees. This means that the isosceles triangle becomes a right triangle and has the maximum area when the included angle is 90 degrees.

Key Concepts

Isosceles Triangle

An isosceles triangle is a type of triangle that has two sides of equal length. This property has several interesting implications.

• Equal Angles: The angles opposite the equal sides are also equal.
• Symmetry: An isosceles triangle is symmetrical along the axis that bisects the vertex angle (the angle between the two equal sides).

In the context of our exercise, the isosceles triangle has two sides, each 10 meters long, forming the characteristic symmetrical shape. The third side of the triangle will vary depending on the included angle, or the angle between the two equal sides. Understanding these properties helps us visualize the dynamics within the triangle as we tackle problems such as calculating the area.

Area of Triangle

Calculating the area of a triangle is an essential skill, especially in trigonometry. For our isosceles triangle, where two sides and the included angle are known, the formula employed is particularly handy: \[ Area = 0.5ab\sin(c) \]Here, \(a\) and \(b\) represent the lengths of the two sides of the triangle, and \(c\) is the included angle. This formula is derived from properties of trigonometry, particularly the sine function, which helps calculate height-related attributes in a triangle.
For the triangle in the exercise:

• The sides are each 10 meters, so substituting these into the formula, we get \( Area = 50\sin(\theta) \).
• This equation illustrates how the area changes as the angle \(\theta\) varies.

Understanding these relationships is crucial in geometrical problems where direct measurements aren't available, and we need to utilize known properties to solve them.

Trigonometric Functions

Trigonometric functions like sine, cosine, and tangent allow us to relate the angles and sides of triangles. In our exercise, the sine function plays a pivotal role. The sine function, \(\sin(\theta)\), represents the ratio of the opposite side to the hypotenuse in a right triangle. Its properties are leveraged to determine areas in non-right triangles, making it incredibly versatile. Given the angle \(\theta\), our area function involves \(50\sin(\theta)\), which shows that the triangle's area increases as the sine of the angle increases, reaching a peak when \(\theta = \frac{\pi}{2}\) radians (or 90 degrees).
Furthermore, the double angle identity \(\sin(\theta) = 2\sin(\theta/2)\cos(\theta/2)\) transforms our area function. This identity is particularly useful for expressing the area in terms of half-angles, broadening our approaches to solve trigonometric problems efficiently. These insights reflect the power and scope of trigonometric functions in understanding shape dimensions throughout various challenges.