Question

review
what is the tangent ratio of $\angle dwx$
$\bigcirc\frac{2}{2\sqrt{2}}$
$\bigcirc\frac{3}{3\sqrt{2}}$
$\bigcirc\frac{4}{4\sqrt{2}}$
$\bigcirc 1$
question id: 018a4b18 - 6ba5 - 7dde - 9725 - b539

Explanation:

Step1: Recall the tangent ratio formula

The tangent of an angle in a right triangle is defined as the ratio of the length of the opposite side to the length of the adjacent side, i.e., \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).

Step2: Identify the sides for \(\angle DWX\)

In right triangle \(WDX\), for \(\angle DWX\):

• The opposite side to \(\angle DWX\) is \(DX\). From the graph, the length of \(DX\) is \(6 - (-2)=8\)? Wait, no, looking at the coordinates: \(W\) is at \((0, - 2)\), \(D\) is at \((8, - 2)\), and \(X\) is at \((8,6)\). So the vertical distance (opposite side) \(DX = 6-(-2)=8\)? Wait, no, the right triangle has legs: horizontal leg \(WD\) and vertical leg \(DX\). Wait, \(WD\) is the horizontal distance from \(W(0, - 2)\) to \(D(8, - 2)\), so \(WD=8 - 0 = 8\). The vertical leg \(DX\) is from \(D(8, - 2)\) to \(X(8,6)\), so \(DX=6-(-2)=8\). Wait, but maybe I made a mistake. Wait, the angle \(\angle DWX\): the adjacent side is \(WD\) (horizontal) and the opposite side is \(DX\) (vertical). Wait, but actually, looking at the slope or the triangle, maybe the triangle is isoceles? Wait, no, let's re - examine. Wait, the coordinates: \(W(0,-2)\), \(D(8,-2)\), \(X(8,6)\). So \(WD\) is horizontal with length \(8\), \(DX\) is vertical with length \(8\). So \(\tan(\angle DWX)=\frac{DX}{WD}=\frac{8}{8} = 1\). Wait, but let's check the smaller triangles. Wait, the red and blue triangles: the slope of \(WX\) is \(\frac{6 - (-2)}{8-0}=\frac{8}{8}=1\), so the angle \(\angle DWX\) has \(\tan\) ratio equal to the slope, which is \(1\). Also, looking at the options, when we simplify \(\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\), \(\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}\), \(\frac{4}{4\sqrt{2}}=\frac{1}{\sqrt{2}}\), but the correct one is \(1\) because the opposite and adjacent sides are equal (both length \(8\)), so \(\tan(\angle DWX)=\frac{\text{opposite}}{\text{adjacent}}=\frac{8}{8}=1\).

Answer:

\(1\)