Question #f1a06 Trigonometry Right Triangles Relating Trigonometric Functions 1 Answer salamat Mar 29, 2017 #sin = sqrt 56/9#, #cos = 5/9#, and #tan = sqrt 56/5# Explanation: #sec = 1/cos# = #9/5#, #= 5/9# based on pyhtogoras theoram for triangle, #sin = sqrt(9^2 - 5^2)/9 = sqrt 56/9#, and #tan = sqrt 56/5#, we also can use #sin/cos# Answer link Related questions What does it mean to find the sign of a trigonometric function and how do you find it? What are the reciprocal identities of trigonometric functions? What are the quotient identities for a trigonometric functions? What are the cofunction identities and reflection properties for trigonometric functions? What is the pythagorean identity? If #sec theta = 4#, how do you use the reciprocal identity to find #cos theta#? How do you find the domain and range of sine, cosine, and tangent? What quadrant does #cot 325^@# lie in and what is the sign? How do you use use quotient identities to explain why the tangent and cotangent function have... How do you show that #1+tan^2 theta = sec ^2 theta#? See all questions in Relating Trigonometric Functions Impact of this question 1200 views around the world You can reuse this answer Creative Commons License