{"id":"cb_practice_pdf:t4-math-m2-q20","source_id":"cb_practice_pdf","source_item_id":"t4-math-m2-q20","external_id":null,"ibn":null,"question_id":null,"program":"SAT","module":"math","domain_code":null,"domain":null,"skill_code":null,"skill":null,"difficulty":null,"score_band":null,"answer_type":"spr","answer_source":"rationale","layout_warning":null,"form_id":"cb_practice_pdf:test-4","module_number":2,"position":20,"stem":"In triangle JKL, cos(K) = 24/51 and angle J is a right angle. What is the value of cos(L)?","stimulus":"","stimulus_html":null,"stem_html":"<p>In triangle JKL, cos(K) = 24/51 and angle J is a right angle. What is the value of cos(L)?</p>","rationale_html":"<p>15 The correct answer is . It’s given that angle J is the right angle in triangle JKL. 17 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 ( )=24 ( )= 8 the hypotenuse. It’s given that cos K . This can be written as cos K . 51 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 ( )2 ( )+2 ( )=2 follows that JK JL KL . Substituting 8n for JK and 17n for KL yields ( )2 ( )+2 ( =)2 2 ( )+2 =2 8n JL 17n . This is equivalent to 64n JL 289n . Subtracting 2 ( )2= 2 64n from each side of this equation yields JL 225n . Taking the square root = ( )=JL of each side of this equation yields JL 15n. Since cos L , it follows that ( )=15n ( )=15 KL cos L , which can be rewritten as cos L . Note that 15/17, .8824, .8823, 17n 17 and 0.882 are examples of ways to enter a correct answer. 50 SAT PRACTICE TEST #4 ANSWER EXPLANATIONS SAT ANSWER EXPLANATIONS nnn MATH: MODULE 2</p>","correct_answer":["15"],"has_media":false,"has_mathml":false,"has_table":false,"source_created_at":null,"source_updated_at":null,"options":[],"attribution":{"source_id":"cb_practice_pdf","source_name":"College Board full-length SAT practice tests (PDF)","rights_holder":"College Board","canonical_url":"https://satsuite.collegeboard.org/practice/practice-tests/paper","retrieved_from":"https://satsuite.collegeboard.org/practice/practice-tests/paper","attribution":"Full-length SAT practice tests 4–11 © College Board, published free at satsuite.collegeboard.org.","disclaimer":"Extracted from PDF layout: equations, figures and tables do not survive intact. Use the original PDFs for anything the text renders poorly. SAT® is a trademark registered by the College Board, which is not affiliated with and does not endorse this project.","license":"College Board copyright. Free to download; not licensed for redistribution.","redistributable":false,"item_count":905}}