Question

find the exact value of cos l in simplest radical form.
triangle with vertices j, k (right angle), l; side jk = √71, side kl = 3, side jl = √80

Explanation:

Step1: Recall the definition of cosine in a right triangle

In a right triangle, the cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse. For angle \( L \) in right triangle \( \triangle JKL \) (right - angled at \( K \)), we need to identify the adjacent side and the hypotenuse with respect to angle \( L \).

Step2: Identify the sides relative to angle \( L \)

• The hypotenuse of a right triangle is the side opposite the right angle. In \( \triangle JKL \), right - angled at \( K \), the hypotenuse is \( JL=\sqrt{80} \).
• The side adjacent to angle \( L \) is \( LK = 3 \), and the side opposite to angle \( L \) is \( JK=\sqrt{71} \).

Step3: Apply the cosine formula

Using the definition of cosine, \( \cos L=\frac{\text{Adjacent side to }L}{\text{Hypotenuse}}=\frac{LK}{JL} \)

We know that \( LK = 3 \) and \( JL=\sqrt{80} \). First, simplify \( \sqrt{80} \): \( \sqrt{80}=\sqrt{16\times5}=4\sqrt{5} \)

So, \( \cos L=\frac{3}{\sqrt{80}}=\frac{3}{4\sqrt{5}} \). To rationalize the denominator, multiply the numerator and denominator by \( \sqrt{5} \):

\( \cos L=\frac{3\times\sqrt{5}}{4\sqrt{5}\times\sqrt{5}}=\frac{3\sqrt{5}}{4\times5}=\frac{3\sqrt{5}}{20} \)

Answer:

\( \frac{3\sqrt{5}}{20} \)