HELP PLEASEE!! What is the value of x in the triangle?
1 answer:
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so just find the "adjacent" side, using the pythagorean theorem, and even though the square root can give you +/-, use the positive one, assuming the angle is in the 1st quadrant
keeping in mind, that the 2nd quadrant is feasible as well, since the opposite side of +2 can also be in the 2nd quadrant
but anyhow, use the I quadrant
Distance formula between 2 points (x1,y1) and (x2,y2)
D=
D=29 and one point is (-5,5) and the other is (k,0)
sub
29=
29=
square both sides
841=
841=
minus 25 both sides
816=
square root both sides don't forget positive and negative roots
+/-4√51=-5-k
add 5 to both sides
5+/-4√51=-k
times -1 both sides
-5+/-4√51=k
the possible values for k are
k=-5+4√51 or k=-5-4√51
D=
D=29 and one point is (-5,5) and the other is (k,0)
sub
29=
29=
square both sides
841=
841=
minus 25 both sides
816=
square root both sides don't forget positive and negative roots
+/-4√51=-5-k
add 5 to both sides
5+/-4√51=-k
times -1 both sides
-5+/-4√51=k
the possible values for k are
k=-5+4√51 or k=-5-4√51