Question

in $\triangle xyz$, $\angle x = 39^{\circ}$ and $\angle y = 21^{\circ}$. $\angle xwz = 90^{\circ}$ and $xy = 5$. find the length of $xw$ to the nearest hundredth.

Explanation:

Step1: Find angle at Z in triangle XYZ

In triangle \( XYZ \), the sum of angles is \( 180^\circ \). So \( \angle Z = 180^\circ - \angle X - \angle Y = 180^\circ - 39^\circ - 21^\circ = 120^\circ \). But maybe better to use trigonometry in right triangles. Wait, \( XWZ \) is right-angled at \( W \), so triangle \( XWZ \) is right-angled at \( W \), and triangle \( YWZ \) is also right-angled at \( W \). Let \( XW = x \), \( WY = 5 - x \), \( ZW = h \).

In triangle \( XWZ \): \( \cos(39^\circ) = \frac{XW}{XZ} \), \( \sin(39^\circ) = \frac{h}{XZ} \)

In triangle \( YWZ \): \( \cos(21^\circ) = \frac{WY}{YZ} \), \( \sin(21^\circ) = \frac{h}{YZ} \)

Also, in triangle \( XYZ \), using the Law of Sines: \( \frac{XY}{\sin(\angle Z)} = \frac{XZ}{\sin(21^\circ)} = \frac{YZ}{\sin(39^\circ)} \)

\( \angle Z = 120^\circ \), so \( \frac{5}{\sin(120^\circ)} = \frac{XZ}{\sin(21^\circ)} \)

\( XZ = \frac{5 \sin(21^\circ)}{\sin(120^\circ)} \)

Then in triangle \( XWZ \), \( \cos(39^\circ) = \frac{XW}{XZ} \), so \( XW = XZ \cos(39^\circ) = \frac{5 \sin(21^\circ) \cos(39^\circ)}{\sin(120^\circ)} \)

Wait, maybe easier: Let's denote \( XW = x \), \( WY = 5 - x \). In right triangle \( XWZ \), \( \tan(39^\circ) = \frac{ZW}{x} \), so \( ZW = x \tan(39^\circ) \). In right triangle \( YWZ \), \( \tan(21^\circ) = \frac{ZW}{5 - x} \), so \( ZW = (5 - x) \tan(21^\circ) \). Therefore, \( x \tan(39^\circ) = (5 - x) \tan(21^\circ) \)

Step2: Solve for x

\( x \tan(39^\circ) + x \tan(21^\circ) = 5 \tan(21^\circ) \)

\( x (\tan(39^\circ) + \tan(21^\circ)) = 5 \tan(21^\circ) \)

\( x = \frac{5 \tan(21^\circ)}{\tan(39^\circ) + \tan(21^\circ)} \)

Calculate \( \tan(39^\circ) \approx 0.8098 \), \( \tan(21^\circ) \approx 0.3839 \)

So numerator: \( 5 \times 0.3839 \approx 1.9195 \)

Denominator: \( 0.8098 + 0.3839 \approx 1.1937 \)

\( x \approx \frac{1.9195}{1.1937} \approx 1.61 \) Wait, no, wait, maybe I mixed up. Wait, in triangle \( XWZ \), angle at X is 39°, so adjacent side is XW, opposite is ZW. In triangle \( YWZ \), angle at Y is 21°, adjacent side is WY, opposite is ZW. So \( ZW = XW \tan(39^\circ) = (XY - XW) \tan(21^\circ) \)

So \( x \tan(39^\circ) = (5 - x) \tan(21^\circ) \)

\( x \tan(39^\circ) + x \tan(21^\circ) = 5 \tan(21^\circ) \)

\( x (\tan(39^\circ) + \tan(21^\circ)) = 5 \tan(21^\circ) \)

\( x = \frac{5 \tan(21^\circ)}{\tan(39^\circ) + \tan(21^\circ)} \)

Wait, but maybe another approach: Use the Law of Sines in triangle \( XYZ \) to find XZ, then use cosine in triangle \( XWZ \).

Law of Sines: \( \frac{XY}{\sin(\angle Z)} = \frac{XZ}{\sin(\angle Y)} \)

\( \angle Z = 180 - 39 - 21 = 120^\circ \), so \( \frac{5}{\sin(120^\circ)} = \frac{XZ}{\sin(21^\circ)} \)

\( XZ = \frac{5 \sin(21^\circ)}{\sin(120^\circ)} \)

\( \sin(120^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660 \)

\( \sin(21^\circ) \approx 0.3584 \)

So \( XZ \approx \frac{5 \times 0.3584}{0.8660} \approx \frac{1.792}{0.8660} \approx 2.07 \)

Then in triangle \( XWZ \), right-angled at W, \( \cos(39^\circ) = \frac{XW}{XZ} \)

\( XW = XZ \cos(39^\circ) \)

\( \cos(39^\circ) \approx 0.7771 \)

So \( XW \approx 2.07 \times 0.7771 \approx 1.61 \). Wait, but let's check with the first method.

First method: \( \tan(39) \approx 0.8098 \), \( \tan(21) \approx 0.3839 \)

\( x = \frac{5 \times 0.3839}{0.8098 + 0.3839} = \frac{1.9195}{1.1937} \approx 1.61 \). Same result.

Wait, but maybe I made a mistake. Wait, let's check the angle at Z. Wait, in triangle XYZ, angles are 39, 21, so 120. Then, the height ZW splits XY into XW and WY. Let's denote XW = x, WY = 5 - x. Then, in triangle XWZ, tan(39) = ZW / x, so ZW = x tan(39). In triangle YWZ, tan(21) = ZW / (5 - x), so ZW = (5 - x) tan(21). Therefore, x tan(39) = (5 - x) tan(21). Solving for x:

x (tan(39) + tan(21)) = 5 tan(21)

x = (5 tan(21)) / (tan(39) + tan(21))

Calculate tan(39) ≈ 0.8098, tan(21) ≈ 0.3839

Numerator: 5 * 0.3839 ≈ 1.9195

Denominator: 0.8098 + 0.3839 ≈ 1.1937

x ≈ 1.9195 / 1.1937 ≈ 1.61

Wait, but let's check with another approach. Let's use the formula for the length of the segment in a triangle with a height. The length of XW can be found by (XY * cos(angle X) * cos(angle Y)) / (cos(angle X) + cos(angle Y))? Wait, no, maybe not. Alternatively, use the formula for the foot of the altitude.

In a triangle, the length of the segment from vertex X to the foot of the altitude (W) is given by \( \frac{XY \cos(\angle X)}{\cos(\angle X) + \cos(\angle Y)} \)? Wait, no, let's derive it.

From the two equations:

\( ZW = XW \tan(\angle X) \)

\( ZW = (XY - XW) \tan(\angle Y) \)

So \( XW \tan(\angle X) = (XY - XW) \tan(\angle Y) \)

\( XW (\tan(\angle X) + \tan(\angle Y)) = XY \tan(\angle Y) \)

\( XW = \frac{XY \tan(\angle Y)}{\tan(\angle X) + \tan(\angle Y)} \)

Plugging in XY = 5, angle X = 39°, angle Y = 21°:

\( XW = \frac{5 \tan(21^\circ)}{\tan(39^\circ) + \tan(21^\circ)} \)

Calculating:

\( \tan(21^\circ) \approx 0.383864035 \)

\( \tan(39^\circ) \approx 0.8097840332 \)

Numerator: 5 * 0.383864035 ≈ 1.919320175

Denominator: 0.8097840332 + 0.383864035 ≈ 1.193648068

XW ≈ 1.919320175 / 1.193648068 ≈ 1.61

Wait, but let's check with Law of Cosines in triangle XYZ to find XZ, then use cosine in right triangle XWZ.

Law of Cosines: \( XZ^2 = XY^2 + YZ^2 - 2 \cdot XY \cdot YZ \cdot \cos(\angle Y) \). No, maybe better to use Law of Sines first.

Law of Sines: \( \frac{XZ}{\sin(21^\circ)} = \frac{XY}{\sin(120^\circ)} \)

So \( XZ = \frac{5 \sin(21^\circ)}{\sin(120^\circ)} \approx \frac{5 * 0.35836795}{0.8660254038} \approx \frac{1.79183975}{0.8660254038} \approx 2.069 \)

Then in right triangle XWZ, \( \cos(39^\circ) = \frac{XW}{XZ} \)

So \( XW = XZ \cos(39^\circ) \approx 2.069 * 0.7771459615 \approx 2.069 * 0.7771 \approx 1.61 \)

Yes, same result. So XW ≈ 1.61

Answer:

\( \boxed{1.61} \)