Question

find the lengths of w, x, y, and z shown in the figure below if xy = 65. round your answers to the nearest tenth. note that the figure is not drawn to scale. figure with triangles, right angles, 19, 20, w, x, y, z labels w = x = y = z =

Explanation:

Step1: Find \( w \) using cosine

In the left right triangle, hypotenuse is 19, angle is \( 41^\circ \), \( w \) is adjacent. So \( \cos(41^\circ)=\frac{w}{19} \), \( w = 19\cos(41^\circ)\approx19\times0.7547\approx14.3 \)

Step2: Find the height (let's call it \( h \)) using sine

\( \sin(41^\circ)=\frac{h}{19} \), \( h = 19\sin(41^\circ)\approx19\times0.6561\approx12.5 \)

Step3: Find \( x \) and \( y \) given \( xy = 65 \) and \( x + y=h + y? \) Wait, no, the height is \( h \), and \( xy = 65 \), also, the height is the same as the vertical segment? Wait, maybe the left triangle's height is \( h \), and \( x + y \) is related? Wait, no, the left triangle has height \( h \), and the right triangle has base 20. Wait, maybe the vertical segment is \( h \), and \( x \) is part, \( y \) is part, with \( xy = 65 \) and \( x + y = h \)? Wait, no, maybe I misread. Wait, the left triangle: hypotenuse 19, angle 41°, right angle at \( w \) and the vertical. So vertical side (height) is \( 19\sin(41^\circ)\approx12.5 \), horizontal side \( w = 19\cos(41^\circ)\approx14.3 \). Now, the right triangle has base 20, and vertical side is \( x + y \)? Wait, no, the problem says \( xy = 65 \). Wait, maybe the vertical segment is split into \( x \) and \( y \), so \( x + y = \) height? Wait, no, the left triangle's vertical side is, say, \( h = 19\sin(41^\circ)\approx12.5 \), but \( xy = 65 \), so maybe \( x \) and \( y \) are such that \( x + y = h \)? No, that can't be, since 12.5*12.5=156.25>65. Wait, maybe the vertical segment is \( y \), and \( x \) is another segment? Wait, maybe the figure has two right triangles: left with hypotenuse 19, angle 41°, right with base 20, and the vertical sides are \( x \) and \( y \) with \( xy = 65 \), and the total vertical length is \( x + y \). Wait, no, maybe the left triangle's vertical side is \( y \), and the right triangle's vertical side is \( x \), with \( xy = 65 \), and the left triangle's vertical side is \( 19\sin(41^\circ)\approx12.5 \), so \( y = 12.5 \)? No, that would make \( x = 65 / 12.5 = 5.2 \), so \( x = 5.2 \), \( y = 12.5 \)? Wait, but then \( x + y = 17.7 \), but the right triangle's vertical side is \( x \), base 20, so \( z = \sqrt{20^2 + x^2} \). Wait, maybe I made a mistake. Let's re-express:

Left triangle:
- Hypotenuse: 19
- Angle: 41°
- Adjacent (w): \( 19\cos(41^\circ) \approx 19 \times 0.7547 \approx 14.3 \)
- Opposite (height, let's call it \( h \)): \( 19\sin(41^\circ) \approx 19 \times 0.6561 \approx 12.5 \)

Now, the right triangle:
- Base: 20
- Vertical side: let's say \( x \) (wait, no, the problem says \( xy = 65 \), so maybe \( x \) and \( y \) are the vertical segments, with \( y \) being the height of the left triangle, so \( y = h \approx 12.5 \), then \( x = 65 / y = 65 / 12.5 = 5.2 \). Wait, but then the vertical side of the right triangle is \( x = 5.2 \)? No, that doesn't make sense. Wait, maybe the vertical segment is \( x \), and \( y \) is another, with \( xy = 65 \), and the left triangle's vertical side is \( y \), so \( y = 19\sin(41^\circ) \approx 12.5 \), so \( x = 65 / 12.5 = 5.2 \). Then the right triangle has vertical side \( x = 5.2 \), base 20, so \( z = \sqrt{20^2 + 5.2^2} \approx \sqrt{400 + 27.04} \approx \sqrt{427.04} \approx 20.7 \). Wait, but that seems off. Alternatively, maybe the total vertical length is \( x + y = 65 \)? No, the problem says \( xy = 65 \). Wait, maybe I misread the problem. Let's start over.

Left triangle: right-angled, hypotenuse 19, angle 41° at the top. So:

- \( w = 19\cos(41^\circ) \approx 19 \times 0.7547 \approx 14.3 \) (adjacent to 41°)
- Vertical side (let's call it \( v \)): \( 19\sin(41^\circ) \approx 19 \times 0.6561 \approx 12.5 \) (opposite to 41°)

Now, the right triangle: right-angled, base 20, vertical side is \( v' \), and \( z \) is hypotenuse. Also, the vertical segment is split into \( x \) and \( y \) with \( xy = 65 \), and \( x + y = v + v' \)? No, the problem says "the figure below" with \( xy = 65 \). Maybe the vertical side of the left triangle is \( y \), and the vertical side of the right triangle is \( x \), so \( xy = 65 \), and the left triangle's vertical side \( y = 19\sin(41^\circ) \approx 12.5 \), so \( x = 65 / 12.5 = 5.2 \). Then:

- \( w = 19\cos(41^\circ) \approx 14.3 \)
- \( x = 5.2 \) (since \( xy = 65 \), \( y = 12.5 \), so \( x = 65 / 12.5 = 5.2 \))
- \( y = 12.5 \) (from left triangle's vertical side)
- \( z = \sqrt{20^2 + x^2} = \sqrt{400 + 5.2^2} = \sqrt{400 + 27.04} = \sqrt{427.04} \approx 20.7 \)

Wait, but let's check \( xy = 5.2 \times 12.5 = 65 \), which matches. So:

Step1: Calculate \( w \)

\( w = 19\cos(41^\circ) \approx 19 \times 0.7547 \approx 14.3 \)

Step2: Calculate \( y \) (vertical side of left triangle)

\( y = 19\sin(41^\circ) \approx 19 \times 0.6561 \approx 12.5 \)

Step3: Calculate \( x \) from \( xy = 65 \)

\( x = \frac{65}{y} = \frac{65}{12.5} = 5.2 \)

Step4: Calculate \( z \) (hypotenuse of right triangle)

\( z = \sqrt{20^2 + x^2} = \sqrt{400 + 5.2^2} = \sqrt{400 + 27.04} = \sqrt{427.04} \approx 20.7 \)

Answer:

\( w \approx 14.3 \), \( x = 5.2 \), \( y \approx 12.5 \), \( z \approx 20.7 \)