Question

question 17 of 26 given the diagram below, what is \\(\cos(45^\circ)\\)? diagram: right triangle with hypotenuse \\(8\sqrt{2}\\), \\(45^\circ\\) angle, right angle; note: \triangle not drawn to scale\ options: a. \\(\frac{1}{\sqrt{2}}\\), b. \\(\sqrt{2}\\), c. \\(4\sqrt{2}\\), d. \\(2\sqrt{2}\\)

Explanation:

Step1: Identify Triangle Type

The triangle is a right - isosceles triangle (one angle is \(90^{\circ}\), another is \(45^{\circ}\), so the third is also \(45^{\circ}\)). In a right - isosceles triangle, the legs are equal. Let the length of each leg be \(x\). Using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\) (where \(a = b=x\) and \(c = 8\sqrt{2}\)), we have \(x^{2}+x^{2}=(8\sqrt{2})^{2}\).
Simplify: \(2x^{2}=64\times2\), then \(x^{2}=64\), so \(x = 8\).

Step2: Recall Cosine Definition

The cosine of an angle in a right triangle is defined as \(\cos(\theta)=\frac{\text{adjacent side}}{\text{hypotenuse}}\). For \(\theta = 45^{\circ}\), the adjacent side to the \(45^{\circ}\) angle has length \(x = 8\) and the hypotenuse has length \(8\sqrt{2}\).

Step3: Calculate \(\cos(45^{\circ})\)

\(\cos(45^{\circ})=\frac{8}{8\sqrt{2}}=\frac{1}{\sqrt{2}}\) (we can also rationalize it as \(\frac{\sqrt{2}}{2}\), but \(\frac{1}{\sqrt{2}}\) is equivalent).

Answer:

A. \(\frac{1}{\sqrt{2}}\)