91Ó°ÊÓ

Inverse tangent, also known as arctangent (denoted as \(\tan^{-1}\)), is a trigonometric function used to find an angle when the tangent of that angle is known. If you have an angle \(\theta\) where \(\tan(\theta) = \frac{a}{3}\), then \(\theta = \tan^{-1} \frac{a}{3}\). This is useful in many applications, including solving triangles and real-world problems involving angles. Understanding this concept helps in converting from the ratio form back to the angle measure. Just remember that the range of \(\tan^{-1}(x)\) is from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

A right triangle is a triangle in which one of the angles is a right angle (90 degrees). It has three sides:

• The hypotenuse (the longest side opposite the right angle)
• The opposite side (the side opposite the angle of interest)
• The adjacent side (the side next to the angle of interest)

Understanding the relationships between these sides is crucial for trigonometry. In our problem, the angle \(\theta\) forms a right triangle where the tangent of \(\theta\) is \(\frac{a}{3}\). This means the opposite side is \(\text{a}\) and the adjacent side is \(\text{3}\).

The Pythagorean theorem is a fundamental principle in geometry that relates the lengths of the sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Mathematically, \[c^2 = a^2 + b^2\]. In our solution:
\(\text{a}\) is one leg,
\(\text{3}\) is the other leg,
\(\text{h}\) is the hypotenuse.
We used this theorem to find the hypotenuse: \(\text{h} = \sqrt{a^2 + 9}\).

The sine function is one of the primary trigonometric functions. It relates the angle to the ratio of the length of the opposite side to the hypotenuse in a right triangle. For an angle \(\theta\), the sine function is defined as \(\text{sin}(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}\). In our exercise, we need to find \(\text{sin}(\theta)\) where \(\theta = \tan^{-1} \frac{a}{3}\). So, we used the definition and the hypotenuse from the Pythagorean theorem to get:
\[ \text{sin}(\theta) = \frac{a}{\sqrt{a^2 + 9}} \]
If needed, you could rationalize this to \[ \frac{a \sqrt{a^2 + 9}}{a^2 + 9} \]. Understanding sine in relation to a right triangle helps solve many trigonometric problems involving angles and side lengths.