Answer:
Unlike the previous problem, this one requires application of the Law of Cosines. You want to find angle Q when you know the lengths of all 3 sides of the triangle.
Law of Cosines: a^2 = b^2 + c^2 - 2bc cos A
Applying that here:
40^2 = 32^2 + 64^2 - 2(32)(64)cos Q
Do the math. Solve for cos Q, and then find Q in degrees and Q in radians.
Step-by-step explanation:
5x+7y=25
5x-2y=-65
subtract both of the equations
5x-5x=0
7y--2y= 7y+2y=9y
25--65=25+65=90
0+9y=90
9y=90
divide both sides by 9 to get y by itself
9y/9=90/9
cross out 9 and 9 which becomes 1*1*y=y
90/9=10
y=10
find x by using the substitution method
5x+7y=25
5x+7(10)=25
5x+70=25
move +70 to the other side
sign changes from +70 to -70
5x+70-70=25-70
5x=-45
divide both sides by 5 to get x by itself
5x/5=-45/5
cross out 5 and 5 then becomes 1*1*x=x
-45/5=-9
x=-9
answer:
(-9,10)