Problem 57 - Precalculus Mathematics for Calculus

Question

A triangle has an area of 16 in \(^{2}\), and two of the sides of the triangle have lengths 5 in. and 7 in. Find the angle included by these two sides.

Step-by-Step Solution

Answer

The angle included by the two sides is approximately 67.4 degrees.

Step 1: Recall the formula for the area of a triangle

The area of a triangle can be calculated using the formula:\[ A = \frac{1}{2}ab\sin(C) \]where \(a\) and \(b\) are the lengths of two sides, and \(C\) is the included angle between them.

Step 2: Plug in the given values into the area formula

We know the area \(A = 16\) square inches, side \(a = 5\) inches, and side \(b = 7\) inches. Replace these values in the formula:\[ 16 = \frac{1}{2} \times 5 \times 7 \times \sin(C) \]

Step 3: Solve for \(\sin(C)\)

Simplify the equation from Step 2:\[ 16 = \frac{35}{2} \cdot \sin(C) \]Multiply both sides by 2 to get:\[ 32 = 35 \cdot \sin(C) \]Divide both sides by 35:\[ \sin(C) = \frac{32}{35} \]

Step 4: Find the angle measure from \(\sin(C)\)

Use the inverse sine function to find the measure of angle \(C\):\[ C = \arcsin\left(\frac{32}{35}\right) \]Calculate this value to find the angle \(C\) in degrees.

Key Concepts

Triangle Area

Understanding the area of a triangle is crucial in geometry. There are several ways to find this area, depending on the information available.
For instance, if you know the base and height of the triangle, the formula is simple:

• Area = \( \frac{1}{2} \times \text{base} \times \text{height} \)

However, if you have two side lengths and the included angle, you can use the sine formula:

• Area = \( \frac{1}{2}ab\sin(C) \)

In the given problem, since we have the lengths of two sides as 5 inches and 7 inches, and the area as 16 square inches, the formula involving sine becomes handy.
This is because knowing the area and two sides directly leads to finding the sine of the included angle using algebra. It’s an elegant method because it circumvents the need for measuring height directly or using Heron’s formula, which requires all three sides.

Sine Function

The sine function is a fundamental concept in trigonometry.
It relates the angle in a right triangle to the ratio of the opposite side over the hypotenuse.

• In symbols: \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \)

In the context of our problem, we use it in a different form called the "sine rule" for triangle area.
This rule is essential when dealing with non-right triangles where two sides and the angle between them are known.
Understanding how to manipulate the sine function in equations is crucial for solving many geometric problems.

• It helps find unknown angles or sides when other measurements are given.

The sine function itself ranges between -1 and 1, which makes it useful in determining possible angle measures.
Since our sine value \( \sin(C) \) ended up between 0 and 1, it confirms that angle \( C \) is viable and acute in the physical scenario of a triangle.

Inverse Sine

The inverse sine function, often written as \( \arcsin \), is used to find angles when their sine values are known.
This is particularly useful in geometry problems where some side lengths and areas are known, but the angles are not.

• Mathematically, if \( \sin(x) = a \), then \( x = \arcsin(a) \)

In the problem, after determining \( \sin(C) = \frac{32}{35} \), the next logical step was to use \( \arcsin \) to find angle \( C \).
While regular sine functions help move from angles to sides or vice versa, \( \arcsin \) does the reverse.
When using the inverse sine function, it's important to remember that it outputs the principal value, which usually is between -90 and 90 degrees.
For our triangle problem, it simply helped derive the angle by utilizing the previously calculated sine value.

• This completes the process and provides a full understanding of the scenario, especially when such triangle problems arise in practice or exams.