Write and solve an inequality for the possible values of \(x\).
Find the third angle of the triangle with angles \(48^\circ\) and \(66^\circ\).
The sum of the angles in a triangle is \(180^\circ\). Let the third angle be \(y\).
Then \(48 + 66 + y = 180\).
\(114 + y = 180\)
\(y = 180 - 114\)
\(y = 66^\circ\)
Find the value of \(x\) using the isosceles triangle theorem.
The triangle with angles \(48^\circ\), \(66^\circ\), and \(66^\circ\) is isosceles. The sides opposite the equal angles are equal.
The triangle with angles \(70^\circ\), \(70^\circ\), and an unknown angle is also isosceles. The sum of the angles in a triangle is \(180^\circ\), so the third angle is \(180 - 70 - 70 = 40^\circ\).
Since the two triangles share a side, the length of the common side is equal.
In the isosceles triangle with angles \(70^\circ\), \(70^\circ\), and \(40^\circ\), the sides opposite the equal angles are equal. The side opposite \(40^\circ\) is given as \(3x - 18\). The side opposite \(66^\circ\) in the other triangle is given as \(2(x+22)\).
Since the two sides are equal, we have \(3x - 18 = 2(x+22)\).
Solve for \(x\).
\(3x - 18 = 2(x + 22)\)
\(3x - 18 = 2x + 44\)
\(3x - 2x = 44 + 18\)
\(x = 62\)
Since we are asked for an inequality, we are looking for possible values of \(x\). Since the side lengths must be positive, we must have \(3x - 18 > 0\) and \(2(x+22) > 0\).
\(3x > 18 \implies x > 6\)
\(2(x+22) > 0 \implies x+22 > 0 \implies x > -22\)
Since \(x>6\) and \(x>-22\), we have \(x>6\).
\(\boxed{x>6}\)
\(\boxed{x>6}\)
Answered
