abelkoontz Posted January 17, 2017 Report Share Posted January 17, 2017 (edited) How do u do 8b) and ci) questions from IB Maths SL M10/5/MATME/SP2/ENG/TZ1/XX+ Edited January 17, 2017 by abelkoontz Reply Link to post Share on other sites More sharing options...
kw0573 Posted January 17, 2017 Report Share Posted January 17, 2017 (edited) b) Sin rule relates two angles with two sides. To limit to one unknown in the equation, we use the angle BAC 30° = angle A. a / sinA = b / sin B. BC / sin (30°) = AC / sin x. AC = (4) sin x / (1/2) = 8 sin x. c)i) Set the two equations equal to each other: 8 sin x = sqrt (41 - 40 cos x) square both sides 64 sin²x = 41 - 40 cos x The straight forward method is to change 64 (sin²x) to 64 (1 - cos²x) using the pythagorean identity 64 - 64 cos² x = 41 - 40 cos x 64 cos²x - 40 cos x - 23 = 0 Solve the quadratic for cos x. Because B is greater than 90° and less than 180°, cos x is between 0 and -1. x≈-0.36354 arccos give angle in 0° to 180°, assure calculator is in degrees, arccos (-0.36354) = 111.32 ° Edited January 17, 2017 by kw0573 Reply Link to post Share on other sites More sharing options...
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