Question

find the value of $x$.
(there is a right triangle with a right angle, a 45° angle, one leg labeled $x$, hypotenuse labeled 12)

Explanation:

Step1: Identify Triangle Type

The triangle is right - angled (one right angle) and has a \(45^{\circ}\) angle. So, it is a 45 - 45 - 90 triangle (isosceles right - triangle), where the legs are equal and the hypotenuse \(h\) is related to the leg \(l\) by \(h = l\sqrt{2}\). But we can also use trigonometry. Let's use cosine function. In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here, \(\theta = 45^{\circ}\), adjacent side is \(x\), and hypotenuse is 12.

Step2: Apply Cosine Formula

We know that \(\cos(45^{\circ})=\frac{x}{12}\). Since \(\cos(45^{\circ})=\frac{\sqrt{2}}{2}\), we can solve for \(x\):

$$x = 12\times\cos(45^{\circ})=12\times\frac{\sqrt{2}}{2}=6\sqrt{2}$$

Or, since it's a 45 - 45 - 90 triangle, the two legs are equal. Let the legs be \(l\) and the hypotenuse be \(h\). We know that \(h = l\sqrt{2}\), so \(l=\frac{h}{\sqrt{2}}\). Substituting \(h = 12\), we get \(l=\frac{12}{\sqrt{2}}=\frac{12\sqrt{2}}{2}=6\sqrt{2}\) (rationalizing the denominator).

Answer:

\(x = 6\sqrt{2}\)