Question

find the lengths of w, x, y, and z shown in the figure below if xy = 63. round your answers to the nearest tenth. note that the figure is not drawn to scale. 22 z x 44° w 12 w = x = y = z =

Explanation:

Step1: Analyze the right triangle with hypotenuse 12 and angle 44°

In the smaller right triangle (with hypotenuse 12 and angle 44°), we can use trigonometric ratios. For angle 44°, the adjacent side to the angle is \( x \) and the opposite side is \( w \), hypotenuse is 12.
- To find \( x \) (adjacent side): \( \cos(44^\circ)=\frac{x}{12} \), so \( x = 12\cos(44^\circ) \). Calculate \( \cos(44^\circ)\approx0.7193 \), so \( x\approx12\times0.7193\approx8.6316 \). But wait, we know \( xy = 63 \), so we can also find \( y \) later. Wait, maybe first find \( w \) (opposite side): \( \sin(44^\circ)=\frac{w}{12} \), so \( w = 12\sin(44^\circ) \). \( \sin(44^\circ)\approx0.6947 \), so \( w\approx12\times0.6947\approx8.3364\approx8.3 \).

Step2: Find \( x \) using \( xy = 63 \) and later? Wait, no, first, in the smaller triangle, we found \( x\approx8.63 \), but let's check. Wait, maybe the two right triangles are similar? Wait, no, the left triangle has a leg 22, and the right triangle has a leg \( w \) and hypotenuse 12. Wait, maybe the angle at the bottom is the same (44°) for both triangles? So the left triangle is also a right triangle with angle 44°, leg 22, and hypotenuse \( z \), and the other leg is \( y \). Wait, and the vertical segments \( x \) and \( y \) add up? Wait, no, the figure shows that \( x \) and \( y \) are two parts of the vertical line, with a right angle between the horizontal segments (22 and \( w \)) and the vertical segments (\( y \) and \( x \)). So the two triangles (left and right) are both right triangles with the same acute angle (44°), so they are similar? Wait, no, the left triangle has horizontal leg 22, vertical leg \( y \), hypotenuse \( z \). The right triangle has horizontal leg \( w \), vertical leg \( x \), hypotenuse 12. And the angle at the bottom is 44° for both. So for the right triangle (with hypotenuse 12):

- \( \cos(44^\circ)=\frac{x}{12} \implies x = 12\cos(44^\circ)\approx12\times0.7193\approx8.63 \)
- \( \sin(44^\circ)=\frac{w}{12} \implies w = 12\sin(44^\circ)\approx12\times0.6947\approx8.34\approx8.3 \)
Now, since \( xy = 63 \), and we can find \( y = \frac{63}{x} \). Then, for the left triangle, which is also a right triangle with angle 44°, horizontal leg 22, vertical leg \( y \), so \( \tan(44^\circ)=\frac{22}{y} \implies y = \frac{22}{\tan(44^\circ)} \). Wait, no, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \), so for angle 44°, opposite is 22, adjacent is \( y \), so \( \tan(44^\circ)=\frac{22}{y} \implies y = \frac{22}{\tan(44^\circ)} \). \( \tan(44^\circ)\approx0.9657 \), so \( y\approx\frac{22}{0.9657}\approx22.78 \). Then, since \( xy = 63 \), \( x = \frac{63}{y}\approx\frac{63}{22.78}\approx2.76 \). Wait, that contradicts the earlier \( x \) from the right triangle. So maybe the angle is in the right triangle, and the left triangle has the same angle, so the two triangles are similar. So the ratio of horizontal legs is equal to the ratio of vertical legs. So \( \frac{22}{w}=\frac{y}{x} \), and \( xy = 63 \), and from the right triangle, \( \frac{w}{x}=\tan(44^\circ) \) (since \( \tan(44^\circ)=\frac{w}{x} \)). So \( w = x\tan(44^\circ) \). Also, \( \frac{22}{w}=\frac{y}{x} \implies \frac{22}{x\tan(44^\circ)}=\frac{y}{x} \implies \frac{22}{\tan(44^\circ)} = y \). Then, since \( xy = 63 \), \( x = \frac{63}{y}=\frac{63\tan(44^\circ)}{22} \). Calculate \( \tan(44^\circ)\approx0.9657 \), so \( x\approx\frac{63\times0.9657}{22}\approx\frac{60.8391}{22}\approx2.765\approx2.8 \). Then \( y = \frac{63}{x}\approx\frac{63}{2.765}\approx22.78\approx22.8 \). Now, for the right triangle, \( w = x\tan(44^\circ)\approx2.765\times0.9657\approx2.67\approx2.7 \)? Wait, no, earlier we thought \( w = 12\sin(44^\circ) \), but that was wrong. Wait, the right triangle has hypotenuse 12, angle 44°, so \( \sin(44^\circ)=\frac{w}{12} \) (opposite over hypotenuse), so \( w = 12\sin(44^\circ)\approx12\times0.6947\approx8.336\approx8.3 \). And \( \cos(44^\circ)=\frac{x}{12} \) (adjacent over hypotenuse), so \( x = 12\cos(44^\circ)\approx12\times0.7193\approx8.6316\approx8.6 \). But then \( xy = 8.6\times22.8\approx196.08 eq63 \). So my initial assumption about the angle is wrong. Wait, maybe the angle is in the left triangle? Let's re-examine the figure. The right triangle has a right angle, a leg of length 12, and angle 44° at the bottom. The left triangle has a right angle, a leg of length 22, and the same angle at the bottom (44°). So the two triangles are similar (AA similarity, both right angles and same acute angle). Therefore, the ratio of corresponding sides is equal. So \( \frac{22}{w}=\frac{y}{x}=\frac{z}{12} \). Also, \( xy = 63 \). Let's denote the ratio as \( k \), so \( 22 = kw \), \( y = kx \), \( z = 12k \). From \( y = kx \), we have \( k = \frac{y}{x} \). Then \( 22 = \frac{y}{x}w \implies 22x = yw \). But \( xy = 63 \implies y = \frac{63}{x} \). Substitute into \( 22x = yw \): \( 22x = \frac{63}{x}w \implies 22x^2 = 63w \). Now, in the right triangle, using Pythagoras: \( x^2 + w^2 = 12^2 = 144 \implies x^2 = 144 - w^2 \). Substitute into \( 22x^2 = 63w \): \( 22(144 - w^2) = 63w \implies 3168 - 22w^2 = 63w \implies 22w^2 + 63w - 3168 = 0 \). Solve this quadratic equation for \( w \). The quadratic formula: \( w = \frac{-63 \pm \sqrt{63^2 + 4\times22\times3168}}{2\times22} \). Calculate discriminant: \( 63^2 = 3969 \), \( 4\times22\times3168 = 88\times3168 = 278784 \), so discriminant \( = 3969 + 278784 = 282753 \). \( \sqrt{282753}\approx531.7 \). So \( w = \frac{-63 + 531.7}{44}\approx\frac{468.7}{44}\approx10.65 \). Wait, that's different from before. So maybe my similarity assumption is wrong. Let's start over.

Right triangle (with hypotenuse 12, angle 44° at the bottom):
- Adjacent side to 44°: \( x \), so \( \cos(44^\circ) = \frac{x}{12} \implies x = 12\cos(44^\circ) \approx 12\times0.7193 \approx 8.63 \)
- Opposite side to 44°: \( w \), so \( \sin(44^\circ) = \frac{w}{12} \implies w = 12\sin(44^\circ) \approx 12\times0.6947 \approx 8.336 \approx 8.3 \)

Left triangle: right triangle, horizontal leg 22, vertical leg \( y \), angle 44° at the bottom. So \( \tan(44^\circ) = \frac{22}{y} \implies y = \frac{22}{\tan(44^\circ)} \approx \frac{22}{0.9657} \approx 22.78 \)

Now, check \( xy = 8.63\times22.78 \approx 196.6 eq 63 \). So the angle is not 44° for both? Wait, maybe the angle in the right triangle is 44°, and the left triangle has a different angle, but the vertical segments \( x \) and \( y \) multiply to 63. So \( x \) is from the right triangle: \( x = 12\cos(44^\circ) \approx 8.63 \), then \( y = 63 / x \approx 63 / 8.63 \approx 7.30 \). Then, the left triangle has horizontal leg 22, vertical leg \( y \approx7.30 \), so hypotenuse \( z = \sqrt{22^2 + y^2} = \sqrt{484 + 53.29} = \sqrt{537.29} \approx 23.18 \approx 23.2 \). And the angle at the bottom for the left triangle would be \( \arctan(22 / y) = \arctan(22 / 7.30) \approx \arctan(3.0137) \approx 71.6° \), which is different from 44°, so the triangles are not similar. So the correct approach is:

1. For the right triangle (hypotenuse 12, angle 44°):
- \( w = 12\sin(44^\circ) \approx 12\times0.6947 \approx 8.3 \)
- \( x = 12\cos(44^\circ) \approx 12\times0.7193 \approx 8.6 \)
2. Given \( xy = 63 \), so \( y = 63 / x \approx 63 / 8.6 \approx 7.3 \)
3. For the left triangle (right triangle, horizontal leg 22, vertical leg \( y \approx7.3 \)):
- \( z = \sqrt{22^2 + y^2} = \sqrt{484 + 53.29} \approx \sqrt{537.29} \approx 23.2 \)

Wait, but this gives \( xy \approx8.6\times7.3\approx62.78\approx63 \), which is close. So let's recalculate with more precision:

- \( \sin(44^\circ) \approx 0.69465837087 \)
- \( w = 12\times0.69465837087 \approx 8.3359 \approx 8.3 \) (rounded to nearest tenth)
- \( \cos(44^\circ) \approx 0.71933980033 \)
- \( x = 12\times0.71933980033 \approx 8.6321 \approx 8.6 \) (rounded to nearest tenth)
- \( y = 63 / 8.6321 \approx 7.298 \approx 7.3 \) (rounded to nearest tenth)
- \( z = \sqrt{22^2 + 7.298^2} = \sqrt{484 + 53.25} = \sqrt{537.25} \approx 23.178 \approx 23.2 \) (rounded to nearest tenth)

Let's check \( xy = 8.6\times7.3 = 62.78 \approx 63 \), which is acceptable.

So:

- \( w \approx 8.3 \)
- \( x \approx 8.6 \)
- \( y \approx 7.3 \)
- \( z \approx 23.2 \)

Wait, but earlier when I thought the angle was the same, it was wrong. The key is that \( xy = 63 \), so we use that to find \( y \) once we have \( x \) from the right triangle.

Answer:

\( w \approx 8.3 \), \( x \approx 8.6 \), \( y \approx 7.3 \), \( z \approx 23.2 \)

(Note: Depending on the exact figure and angle placement, there might be a different interpretation, but based on the given information, this is the solution.)