given right triangle jkl, what is the value of cos(l)?\no $\frac{5}{13}$\no $\frac{5}{12}$\no…

given right triangle jkl, what is the value of cos(l)?\no $\frac{5}{13}$\no $\frac{5}{12}$\no $\frac{12}{13}$\no $\frac{12}{5}$

given right triangle jkl, what is the value of cos(l)?\no $\frac{5}{13}$\no $\frac{5}{12}$\no $\frac{12}{13}$\no $\frac{12}{5}$

Answer

Explanation:

Step1: Find the hypotenuse

Use the Pythagorean theorem (a^{2}+b^{2}=c^{2}), where (a = 5), (b=12). So (c=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13).

Step2: Recall cosine - definition

The cosine of an angle in a right - triangle is defined as (\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}). For (\cos(L)), the adjacent side to angle (L) is (KL = 5) and the hypotenuse is (JL=13). So (\cos(L)=\frac{5}{13}).

Answer:

(\frac{5}{13})