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Geometry and Trigonometry / Right triangles and trigonometry Difficulty: Hard

In triangle $J K L$ , $\cos (K) = \frac{24}{51}$ and angle $J$ is a right angle. What is the value of $\cos (L)$ ?

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Explanation

The correct answer is $StartFraction 15 Over 17 EndFraction$ . It's given that angle $J$ is the right angle in triangle $J K L$ . Therefore, the acute angles of triangle $J K L$ are angle $K$ and angle $L$ . The hypotenuse of a right triangle is the side opposite its right angle. Therefore, the hypotenuse of triangle $J K L$ is side $K L$ . The cosine of an acute angle in a right triangle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. It's given that $\cos (K) = \frac{24}{51}$ . This can be written as $\cos (K) = \frac{8}{17}$ . Since the cosine of angle $K$ is a ratio, it follows that the length of the side adjacent to angle $K$ is $8 n$ and the length of the hypotenuse is $17 n$ , where $n$ is a constant. Therefore, $upper J upper K equals 8 n$ and $upper K upper L equals 17 n$ . The Pythagorean theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. For triangle $J K L$ , it follows that ${(J K)}^{2} + {(J L)}^{2} = {(K L)}^{2}$ . Substituting $8 n$ for $J K$ and $17 n$ for $K L$ yields ${(8 n)}^{2} + {(J L)}^{2} = {(17 n)}^{2}$ . This is equivalent to $64 n^{2} + {(J L)}^{2} = 289 n^{2}$ . Subtracting $64 n squared$ from each side of this equation yields $left parenthesis upper J upper L right parenthesis squared equals 225 n squared$ . Taking the square root of each side of this equation yields $upper J upper L equals 15 n$ . Since $\cos (L) = \frac{J L}{K L}$ , it follows that $\cos (L) = \frac{15 n}{17 n}$ , which can be rewritten as $\cos (L) = \frac{15}{17}$ . Note that 15/17, .8824, .8823, and 0.882 are examples of ways to enter a correct answer.