OG详解-OG4 数学2 Q20

The correct answer is 15​17​. It’s given that angle J is the right angle in triangle JKL. Therefore, the acute angles of triangle JKL are angle K and angle L. The hypotenuse of a right triangle is the side opposite its right angle. Therefore, the hypotenuse of triangle JKL is side KL. 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)= 24​51​ . This can be written as cos (K)= 8​17​.  Since the cosine of angle K is a ratio, it follows that the length of the side adjacent to angle K is 8n and the length of the hypotenuse is 17n , where n is a constant. Therefore, JK = 8n and KL = 17n. 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  JKL, it follows that (JK)² + (JL)²  = (KL)². Substituting 8n for JK  and 17n for KL yields (8n)² + (JL)²  = (17n)² . This is equivalent to 64n²  + (JL)²  = 289n² . Subtracting 64n²  from each side of this equation yields  (JL)²  = 225n². Taking the square root of each side of this equation yields JL = 15n. Since cos(L)= JLKL​​, it follows that cos (L)= 15n17n​​, which can be rewritten as cos(L)= 15​17​. Note that 15/17, .8824, .8823, and 0.882 are examples of ways to enter a correct answer.