Question

15 mark for review in triangle xyz, angle y is a right angle, the measure of angle z is 33°, and the length of \\(\overline{yz}\\) is 26 units. if the area, in square units, of triangle xyz can be represented by the expression \\(k\tan 33°\\), where \\(k\\) is a constant, what is the value of \\(k\\)?

Explanation:

Step1: Identify triangle sides

In right triangle \( XYZ \), \( \angle Y = 90^\circ \), \( \angle Z = 33^\circ \), \( YZ = 26 \). Let \( XY = a \), \( YZ = b = 26 \), \( XY \) is opposite \( \angle Z \), \( YZ \) is adjacent.

Step2: Use tangent ratio

\( \tan 33^\circ=\frac{XY}{YZ}=\frac{a}{26} \), so \( a = 26\tan 33^\circ \).

Step3: Calculate area

Area of triangle \(=\frac{1}{2}\times YZ\times XY=\frac{1}{2}\times26\times a \). Substitute \( a = 26\tan 33^\circ \):
\( \text{Area}=\frac{1}{2}\times26\times26\tan 33^\circ = 338\tan 33^\circ \). Wait, no, wait: Wait, \( YZ = 26 \), \( XY \) is height, \( YZ \) is base? Wait, no, \( \angle Z = 33^\circ \), so \( \tan 33^\circ=\frac{XY}{YZ}\), so \( XY = YZ\tan 33^\circ = 26\tan 33^\circ \). Then area is \( \frac{1}{2}\times YZ\times XY=\frac{1}{2}\times26\times26\tan 33^\circ \)? No, wait, no: Wait, \( YZ \) is adjacent, \( XY \) is opposite. So base is \( YZ = 26 \), height is \( XY \). So area \(=\frac{1}{2}\times26\times XY \). And \( XY = YZ\tan 33^\circ = 26\tan 33^\circ \). So area \(=\frac{1}{2}\times26\times26\tan 33^\circ \)? Wait, no, that can't be. Wait, the problem says area is \( k\tan 33^\circ \). So let's re - express:

Wait, area \(=\frac{1}{2}\times YZ\times XY \). \( XY = YZ\tan 33^\circ = 26\tan 33^\circ \). So area \(=\frac{1}{2}\times26\times26\tan 33^\circ \)? No, that would be \( 338\tan 33^\circ \), but the problem says \( k\tan 33^\circ \). Wait, maybe I mixed up the sides. Wait, \( \angle Z = 33^\circ \), so the sides: \( YZ \) is adjacent to \( \angle Z \), \( XY \) is opposite, \( XZ \) is hypotenuse. So \( \tan 33^\circ=\frac{XY}{YZ}\), so \( XY = YZ\tan 33^\circ = 26\tan 33^\circ \). Then area is \( \frac{1}{2}\times YZ\times XY=\frac{1}{2}\times26\times26\tan 33^\circ \)? No, that's not matching. Wait, no, the problem says the area is \( k\tan 33^\circ \). So let's check again. Wait, maybe \( YZ \) is one leg, \( XY \) is the other leg. Wait, area \(=\frac{1}{2}\times YZ\times XY \). Let \( YZ = 26 \), \( XY = x \). Then \( \tan 33^\circ=\frac{XY}{YZ}=\frac{x}{26}\), so \( x = 26\tan 33^\circ \). Then area \(=\frac{1}{2}\times26\times x=\frac{1}{2}\times26\times26\tan 33^\circ = 338\tan 33^\circ \). But the problem says area is \( k\tan 33^\circ \), so \( k = 338 \)? Wait, no, that can't be. Wait, maybe I made a mistake in the legs. Wait, \( \angle Y = 90^\circ \), so legs are \( XY \) and \( YZ \). So \( YZ = 26 \), \( XY \) is the other leg. Then \( \tan 33^\circ=\frac{XY}{YZ}\), so \( XY = 26\tan 33^\circ \). Then area is \( \frac{1}{2}\times YZ\times XY=\frac{1}{2}\times26\times26\tan 33^\circ = 338\tan 33^\circ \). So \( k = 338 \)? Wait, but let's check again. Wait, the problem says "the area... can be represented by the expression \( k\tan 33^\circ \)". So if we have \( \frac{1}{2}\times26\times XY \), and \( XY = 26\tan 33^\circ \), then area is \( \frac{1}{2}\times26\times26\tan 33^\circ = 338\tan 33^\circ \), so \( k = 338 \). Wait, but maybe I messed up the sides. Wait, \( \angle Z = 33^\circ \), so the sides: \( YZ \) is adjacent to \( \angle Z \), \( XY \) is opposite, \( XZ \) is hypotenuse. So yes, \( \tan\theta=\frac{\text{opposite}}{\text{adjacent}}=\frac{XY}{YZ} \). So \( XY = YZ\tan 33^\circ = 26\tan 33^\circ \). Then area is \( \frac{1}{2}\times YZ\times XY=\frac{1}{2}\times26\times26\tan 33^\circ = 338\tan 33^\circ \). So \( k = 338 \).

Wait, no, wait a second. Wait, the problem says "the area... can be represented by the expression \( k\tan 33^\circ \)". Let's re - do:

Let me denote:

- Right triangle \( XYZ \), right - angled at \( Y \). So \( \angle Y = 90^\circ \), \( \angle Z = 33^\circ \), \( YZ = 26 \).

- Let \( XY = a \) (opposite \( \angle Z \)), \( YZ = b = 26 \) (adjacent to \( \angle Z \)), \( XZ = c \) (hypotenuse).

- \( \tan(33^\circ)=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b}=\frac{a}{26}\), so \( a = 26\tan(33^\circ) \).

- Area of triangle \( A=\frac{1}{2}\times b\times a=\frac{1}{2}\times26\times26\tan(33^\circ)=338\tan(33^\circ) \).

- The problem states that \( A = k\tan(33^\circ) \). So by comparing, \( k = 338 \).

Answer:

\( 338 \)