Question

find: \\(\cos z\\)
find: \\(\sin n\\)

Explanation:

First Sub - Question: Find \(\cos Z\)

Step 1: Identify the triangle type and sides

We have a right - triangle \(XYZ\) with \(\angle X = 90^{\circ}\), \(XY = 24\), \(XZ=18\). First, we need to find the length of the hypotenuse \(YZ\) using the Pythagorean theorem \(c=\sqrt{a^{2}+b^{2}}\), where \(a = 24\), \(b = 18\) and \(c\) is the hypotenuse.
\(YZ=\sqrt{24^{2}+18^{2}}=\sqrt{576 + 324}=\sqrt{900}=30\)

Step 2: Recall the definition of cosine

The cosine of an angle in a right - triangle is defined as \(\cos\theta=\frac{\text{adjacent side to } \theta}{\text{hypotenuse}}\). For \(\angle Z\), the adjacent side is \(XZ = 18\) and the hypotenuse is \(YZ=30\).
\(\cos Z=\frac{XZ}{YZ}=\frac{18}{30}=\frac{3}{5}\)

Second Sub - Question: Find \(\sin N\)

Step 1: Identify the triangle type and sides

We have a right - triangle \(LMN\) with \(\angle M = 90^{\circ}\), \(MN = 15\), \(LN=39\). First, we need to find the length of the side \(LM\) using the Pythagorean theorem \(a=\sqrt{c^{2}-b^{2}}\), where \(c = 39\) (hypotenuse), \(b = 15\) (one leg) and \(a\) is the other leg.
\(LM=\sqrt{39^{2}-15^{2}}=\sqrt{(39 + 15)(39 - 15)}=\sqrt{54\times24}=\sqrt{1296}=36\)

Step 2: Recall the definition of sine

The sine of an angle in a right - triangle is defined as \(\sin\theta=\frac{\text{opposite side to } \theta}{\text{hypotenuse}}\). For \(\angle N\), the opposite side is \(LM = 36\) and the hypotenuse is \(LN = 39\).
\(\sin N=\frac{LM}{LN}=\frac{36}{39}=\frac{12}{13}\)

Answer:

s:

• \(\cos Z=\frac{3}{5}\)
• \(\sin N=\frac{12}{13}\)