Question

find the length of side w in the picture, below.
triangle image with 45° angle, right angle, base \\(\frac{\sqrt{2}}{2}\\), hypotenuse w
w =
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Explanation:

Step1: Identify triangle type

The triangle is a right - isosceles triangle (one angle is \(45^{\circ}\), right - angled), so the adjacent side to \(45^{\circ}\) and the opposite side are equal, and we can use trigonometric ratios. Let's use the cosine function. The cosine of an angle \(\theta\) in a right triangle is defined as \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). Here, \(\theta = 45^{\circ}\), the adjacent side to \(45^{\circ}\) is \(\frac{\sqrt{2}}{2}\), and the hypotenuse is \(w\).

Step2: Recall the value of \(\cos45^{\circ}\)

We know that \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\). From the cosine formula \(\cos45^{\circ}=\frac{\frac{\sqrt{2}}{2}}{w}\).

Step3: Solve for \(w\)

Since \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), we have the equation \(\frac{\sqrt{2}}{2}=\frac{\frac{\sqrt{2}}{2}}{w}\). Cross - multiplying gives us \(\frac{\sqrt{2}}{2}\times w=\frac{\sqrt{2}}{2}\). Dividing both sides by \(\frac{\sqrt{2}}{2}\) (or multiplying both sides by \(\frac{2}{\sqrt{2}}\)), we get \(w = 1\). We can also use the fact that in a \(45 - 45-90\) triangle, the hypotenuse \(h\) is related to the leg \(l\) by \(h = l\sqrt{2}\). Wait, no, in a \(45 - 45-90\) triangle, if the leg length is \(l\), the hypotenuse is \(l\sqrt{2}\). Wait, let's re - check. Wait, the leg length here: if we consider the adjacent side (leg) as \(l=\frac{\sqrt{2}}{2}\), then hypotenuse \(w\). Wait, maybe using the cosine was a bit confusing. Let's use the definition of cosine correctly. \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), so \(\cos45^{\circ}=\frac{\text{adjacent}}{w}\), adjacent is \(\frac{\sqrt{2}}{2}\), \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), so \(\frac{\sqrt{2}}{2}=\frac{\frac{\sqrt{2}}{2}}{w}\), then \(w = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1\). Alternatively, since it's a \(45 - 45-90\) triangle, the two legs are equal. Let the length of each leg be \(l\). Then, by Pythagoras theorem, \(w^{2}=l^{2}+l^{2}=2l^{2}\), so \(w = l\sqrt{2}\). Here, \(l=\frac{\sqrt{2}}{2}\), so \(w=\frac{\sqrt{2}}{2}\times\sqrt{2}=\frac{2}{2} = 1\).

Answer:

\(1\)