Question

find the lengths of w, x, y, and z shown in the figure below if xy=62. round your answers to the nearest tenth. note that the figure is not drawn to scale.

Explanation:

Step1: Analyze the right triangle with hypotenuse 11 and angle 40°

In the upper right triangle, we can use trigonometric ratios. For angle \(40^\circ\), \(\sin(40^\circ)=\frac{w}{11}\) and \(\cos(40^\circ)=\frac{x}{11}\).
First, calculate \(w\):
\(w = 11\times\sin(40^\circ)\)
\(\sin(40^\circ)\approx0.6428\), so \(w\approx11\times0.6428\approx7.1\)
Then, calculate \(x\):
\(x = 11\times\cos(40^\circ)\)
\(\cos(40^\circ)\approx0.7660\), so \(x\approx11\times0.7660\approx8.4\) But wait, the problem says \(xy = 62\), so maybe we need to use the other triangle. Wait, the lower right triangle has one leg 21, and the other leg \(y\), and the same \(x\) (since it's the adjacent side for the angle, and the two triangles share the same angle's adjacent side? Wait, maybe the two triangles are similar? Wait, no, the upper triangle has hypotenuse 11, angle 40°, right angle. The lower triangle has one leg 21, and the other leg \(y\), and the same \(x\) (the horizontal segment). Wait, the problem states \(xy = 62\). Wait, maybe I misread. Let's re - examine.

Wait, the upper triangle: right - angled, hypotenuse 11, angle \(40^\circ\) at the vertex. So the side opposite \(40^\circ\) is \(w\), adjacent is \(x\). So \(w = 11\sin(40^\circ)\), \(x = 11\cos(40^\circ)\). Then the lower triangle: right - angled, one leg is 21, the other leg is \(y\), and the hypotenuse is \(z\), and the horizontal leg is \(x\) (since the two right - angled triangles share the same horizontal segment \(x\)). And we know that \(xy=62\), so \(y=\frac{62}{x}\)

Wait, maybe the upper triangle: let's recalculate \(x\) correctly. Wait, if we consider the lower triangle: area? No, \(xy = 62\), and the lower triangle has legs \(x\) and 21? No, the lower triangle has legs \(y\) and 21? Wait, the diagram: the upper triangle has a right angle, hypotenuse 11, angle 40°, vertical side \(w\), horizontal side \(x\). The lower triangle has a right angle, vertical side 21, horizontal side \(y\), and the same horizontal side \(x\) (so the horizontal segment is \(x\), and the lower triangle's horizontal side is \(y\)? No, maybe the two triangles are such that the horizontal segment is \(x\), and the upper triangle's horizontal side is \(x\), and the lower triangle's horizontal side is \(y\), and \(xy = 62\). And the lower triangle's vertical side is 21, so the hypotenuse \(z=\sqrt{x^{2}+21^{2}}\) or \(z=\sqrt{y^{2}+21^{2}}\)? Wait, no, the lower triangle: right - angled, legs \(x\) and 21? No, the right angle is at the left, so the legs are \(y\) (horizontal) and 21 (vertical), and the hypotenuse \(z\), and the other leg (horizontal) is \(x\)? Wait, I think I made a mistake in the initial approach.

Let's start over.

1. For the upper right - triangle (angle \(40^\circ\), hypotenuse \(c = 11\)):
- \(\sin(40^{\circ})=\frac{w}{11}\Rightarrow w = 11\sin(40^{\circ})\)
- \(\sin(40^{\circ})\approx0.6428\), so \(w\approx11\times0.6428 = 7.1\)
- \(\cos(40^{\circ})=\frac{x}{11}\Rightarrow x = 11\cos(40^{\circ})\)
- \(\cos(40^{\circ})\approx0.7660\), so \(x\approx11\times0.7660 = 8.4\) But if \(xy = 62\), then \(y=\frac{62}{x}\approx\frac{62}{8.4}\approx7.4\)
2. For the lower right - triangle (legs \(y\) and 21, hypotenuse \(z\)):
- Using the Pythagorean theorem, \(z=\sqrt{y^{2}+21^{2}}\)
- If \(y\approx7.4\), then \(z=\sqrt{7.4^{2}+21^{2}}=\sqrt{54.76 + 441}=\sqrt{495.76}\approx22.3\)

Wait, maybe the correct approach is:

We know that in the upper triangle, \(x = 11\cos(40^{\circ})\approx8.4\), then \(y=\frac{62}{x}=\frac{62}{8.4}\approx7.4\)

In the upper triangle, \(w = 11\sin(40^{\circ})\approx7.1\)

In the lower triangle, \(z=\sqrt{x^{2}+21^{2}}\)? No, wait, the lower triangle has legs \(y\) and 21, so \(z=\sqrt{y^{2}+21^{2}}\). If \(y = \frac{62}{x}\) and \(x = 11\cos(40^{\circ})\)

Let's calculate \(x\) precisely: \(\cos(40^{\circ})\approx0.7660444431\), so \(x = 11\times0.7660444431\approx8.4265\)

Then \(y=\frac{62}{x}=\frac{62}{8.4265}\approx7.36\approx7.4\)

\(w = 11\sin(40^{\circ})\), \(\sin(40^{\circ})\approx0.6427876097\), so \(w = 11\times0.6427876097\approx7.0707\approx7.1\)

For \(z\), in the lower right - triangle, the legs are \(y\) and 21, so \(z=\sqrt{y^{2}+21^{2}}\). Substitute \(y\approx7.4\):

\(z=\sqrt{7.4^{2}+21^{2}}=\sqrt{54.76 + 441}=\sqrt{495.76}\approx22.3\)

Step2: Recalculate with correct relations

Wait, maybe the two triangles are such that the upper triangle has hypotenuse 11, angle \(40^{\circ}\), so \(w = 11\sin(40^{\circ})\approx7.1\), \(x = 11\cos(40^{\circ})\approx8.4\)

Since \(xy = 62\), then \(y=\frac{62}{x}=\frac{62}{8.4}\approx7.4\)

For the lower triangle, using Pythagoras: \(z=\sqrt{x^{2}+21^{2}}\)? No, the lower triangle has legs \(y\) and 21, so \(z=\sqrt{y^{2}+21^{2}}\)

\(y=\frac{62}{x}\), and \(x = 11\cos(40^{\circ})\)

\(x = 11\times\cos(40^{\circ})\approx11\times0.7660 = 8.426\)

\(y=\frac{62}{8.426}\approx7.36\)

\(z=\sqrt{y^{2}+21^{2}}=\sqrt{7.36^{2}+21^{2}}=\sqrt{54.1696 + 441}=\sqrt{495.1696}\approx22.25\approx22.3\)

Answer:

\(w\approx7.1\), \(x\approx8.4\), \(y\approx7.4\), \(z\approx22.3\) (Rounded to the nearest tenth)