A side of the triangle below has been extended to form an exterior angle of 144°. Find the value of x.
Since the exterior angle is equal to the sum of the two remote interior angles, we must have $x+30^\circ = 144^\circ$. Subtracting $30^\circ$ from both sides gives $x=\boxed{114^\circ}$. [asy]
pair A,B,C;
A=(0,0);
B=(4,0);
C=(1,1.732);
draw(A--B--C--cycle);
draw(C--(-1,0));
label("$x$",(0.5,0.2));
label("$60^{\circ}$",(-0.5,1.6));
label("$60^{\circ}$",(2.5,1.6));
label("$30^{\circ}$",(0.5,-0.3));
label("$144^{\circ}$",(1,-0.7));
dot((0,0));
dot((4,0));
dot((1,1.732));
draw(rightanglemark((1,1.732),(4,0),(0,0),8));
[/asy]
nuh-uh.
Apologies for the mistake. Let's try again.
Since the exterior angle is equal to the sum of the two remote interior angles, we have $x+30^\circ = 180^\circ - 144^\circ$. Simplifying the right side gives $x+30^\circ = 36^\circ$. Subtracting $30^\circ$ from both sides gives $x=\boxed{6^\circ}$. [asy]
pair A,B,C;
A=(0,0);
B=(4,0);
C=(1,1.732);
draw(A--B--C--cycle);
draw(C--(-1,0));
label("$x$",(0.5,0.2));
label("$60^\circ$",(-0.5,1.6));
label("$60^\circ$",(2.5,1.6));
label("$30^\circ$",(0.5,-0.3));
label("$144^\circ$",(3,0.5));
dot((0,0));
dot((4,0));
dot((1,1.732));
draw(rightanglemark((1,1.732),(4,0),(0,0),8));
[/asy]
whaaa
I apologize for the mistake. Let's reevaluate the problem.
Let the extended side of the triangle have length $y$. Since the exterior angle is equal to the sum of the two remote interior angles, we have $x + 60^\circ = 144^\circ$. Simplifying gives $x= 84^\circ$.
Then, in the triangle below, we have an angle of $x= 84^\circ$, an angle of $60^\circ$, and the exterior angle of $144^\circ$. Since the angles in a triangle sum to $180^\circ$, we can find the remaining angle by subtracting the sum of the other two angles from $180^\circ$.
The remaining angle is $180^\circ - (x + 60^\circ) = 36^\circ$.
Since the exterior angle is equal to the sum of the remote interior angles, we can set up the equation $y = 144^\circ - 36^\circ = 108^\circ$.
Therefore, $x = \boxed{84^\circ}$. [asy]
pair A,B,C;
A=(0,0);
B=(4,0);
C=(1,1.732);
draw(A--B--C--cycle);
draw(C--(-1,0));
label("$x$",(0.3,0.2));
label("$60^\circ$",(-0.5,1.6));
label("$60^\circ$",(2.5,1.6));
label("$144^\circ$",(1.4,0.5));
dot((0,0));
dot((4,0));
dot((1,1.732));
draw(rightanglemark((1,1.732),(4,0),(0,0),8));
label("$y$", (1.7,-0.7));
[/asy]
To find the value of x, we need to use the concept of exterior angles of a triangle. An exterior angle of a triangle is formed when one of the sides is extended. This exterior angle is equal to the sum of the two opposite interior angles.
In this case, the exterior angle measures 144°. Now, we need to identify the two opposite interior angles. Let's label the triangle as ABC, where AB, BC, and AC are the sides of the triangle.
The exterior angle is formed by extending side AC, so the interior angles adjacent to it are ∠ABC and ∠BCA. We are given that the measure of the exterior angle is 144°.
Since the sum of the exterior angle and the adjacent interior angles is 180°, we can write an equation:
∠ABC + ∠BCA + Exterior Angle = 180°
Substituting the known values:
∠ABC + ∠BCA + 144° = 180°
Now, we need to solve for x. To do that, we need to apply the properties of angles in a triangle.
A triangle has three interior angles, and their sum is always 180°. From the equation above, we know that:
∠ABC + ∠BCA + 144° = 180°
Since the sum of the interior angles of a triangle is always 180°, we can write:
∠ABC + ∠BCA = 180° - 144°
Simplifying:
∠ABC + ∠BCA = 36°
To find the value of x, we need to determine the angle measures ∠ABC and ∠BCA. However, without further information about the triangle, we cannot determine the exact values of these angles. Therefore, we cannot find the value of x with the given information.