Question

in △fgh,\\(\overline{hf}\cong\overline{gh}\\) and \\(m\angle h = 155^{circ}\\). find \\(m\angle g\\).

Explanation:

Step1: Recall the property of isosceles triangle

In \(\triangle FGH\), since \(\overline{HF}\cong\overline{GH}\), then \(\angle F=\angle G\) (base - angles of an isosceles triangle are equal).

Step2: Use the angle - sum property of a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(m\angle F + m\angle G+m\angle H=180^{\circ}\). Let \(m\angle G = x\), then \(m\angle F=x\) (because \(\angle F\cong\angle G\)).
Substitute into the angle - sum formula: \(x + x+155^{\circ}=180^{\circ}\).
Combine like terms: \(2x+155^{\circ}=180^{\circ}\).
Subtract \(155^{\circ}\) from both sides: \(2x=180^{\circ}- 155^{\circ}\).
So, \(2x = 25^{\circ}\).
Divide both sides by 2: \(x=\frac{25^{\circ}}{2}=12.5^{\circ}\).

Answer:

\(m\angle G = 12.5^{\circ}\)