Ineed i’ve been stuck on this for a while now and i can’t get passed it can someone

Atriangular field has sides of lengths 21, 39, and 41 yd. find the largest angle. (round your answer to the nearest degree.)

Worth 100 points need the answers asap first row -x^2 2x^2 (x/2)^2 x^2 x is less than 2 x is greater than 2 x is less than or equal to 2 x is greater than or equal to 2 second row -5 -5/2 4 5 •2 is less than x& x is less than 4 •2 is less than or equal to x & x is less than or equal to 4 •2 is less than or equal to x& x is less than 4 •2 is less than x& x is less than or equal to 4

I've been working on this for a few days and i just don't understand, it's due in a few hours. you. the direction of a vector is defined as the angle of the vector in relation to a horizontal line. as a standard, this angle is measured counterclockwise from the positive x-axis. the direction or angle of v in the diagram is α.part a: how can you use trigonometric ratios to calculate the direction α of a general vector v = < x, y> similar to the diagram? part bsuppose that vector v lies in quadrant ii, quadrant iii, or quadrant iv. how can you use trigonometric ratios to calculate the direction (i. e., angle) of the vector in each of these quadrants with respect to the positive x-axis? the angle between the vector and the positive x-axis will be greater than 90 degrees in each case. part cnow try a numerical problem. what is the direction of the vector w = < -1, 6 > ?