I need to know how I get 65 degrees out of sin(.909) because sin of (.909)=.79 and sin^-1=1.14 and I do not have arcsin on my calculator 9:38AB =c = 12 ; AC =b; BC= a = 5Applying pythagorous theorem in the given rightangle triangle(Hypotenuse)? = (Base)2 + (Perpendicular)?c2 = a² + b²122 = 52 + b²b2 = 144 -25 = 119%3D%3Dthen b = V119Step2b)According to the given figuresinB = Perpendicular / Hypotenuse= b/c= V119/12So,sinB = V119/ 12= 10.90/12 = 0.9090%3DB= sin'(0.9090) = 65.36ºAnswer%3D 10:13AAA bartleby.comUpen in the Bartieby app= bartlebyQ&AMath / Trigonometry / Q&A Library / The Sin^-1(.9090) does ..The Sin^-1(.9090) does not equal to 65 de...Get live help whenever youTry bartleby->tutor todayneed from online tutors!Expert AnswerStep 1To findarcsin(.9090)Step 2arcsin(.9090) = 65.36°It is correctWas this solution helpful?Privacy - Terms

Name: I need to know how I get 65 degrees out of sin(.909) because sin of (.
Uploaded: 2024-03-20T06:46:55.000Z
Channel: Bartleby
Description: I need to know how I get 65 degrees out of sin(.909) because sin of (.909)=.79 and sin^-1=1.14 and I do not have arcsin on my calculator 9:38AB =c = 12 ; AC =b; BC= a = 5Applying pythagorous theorem in the given rightangle triangle(Hypotenuse)? = (Ba