Solve using cosine law only

Please please help this is due tomorrow morning

Hints:

Step 1: Use Sine law the the angle B.

a/Sin A=b/SinB=c/SinC

24/Sin61^=12/SinB (accross multiply)

SinB=12Sin61^/24

SinB=0.437309853 then convert to degree =25.9 degree( depends on question rounded how many decimal) Read the question it must notes there.

--> Angle C=180-61-25.9=93.1 degree.

Step 2: Use Cosine law to calculate.

c^2=a^2+b^2−2ab⋅cosC

c^2=24^2+12^2-2(24)(12)cos93.1

c^2=720-576(-0.054078812)

c^2=720+31.11494

c^2=751.1494

c=27.4

(Square root take exponent off)

Step-by-step explanation:

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.