Question

find the cosine of $\angle j$.

write your answer in simplified, rationalized form. do not round.

$\cos(j)=\square$

Explanation:

Step1: Recall the cosine formula

In a right - triangle, \(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\angle J\), first find the length of \(IJ\) using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 7\), \(c=\sqrt{66}\), and \(b = IJ\). So \(IJ=\sqrt{(\sqrt{66})^{2}-7^{2}}=\sqrt{66 - 49}=\sqrt{17}\).

Step2: Calculate \(\cos(J)\)

\(\cos(J)=\frac{IJ}{JK}\), since \(IJ=\sqrt{17}\) and \(JK = \sqrt{66}\), then \(\cos(J)=\frac{\sqrt{17}}{\sqrt{66}}=\frac{\sqrt{17}\times\sqrt{66}}{\sqrt{66}\times\sqrt{66}}=\frac{\sqrt{1122}}{66}\).

Answer:

\(\frac{\sqrt{1122}}{66}\)