Question

given △abc with altitudes \\(\overline{ad}\\), \\(\overline{be}\\), and \\(\overline{cf}\\) intersecting at orthocenter point g.

if \\( m\angle bgf = (7x - 4)\degree \\) and \\( m\angle fbg = (5x - 14)\degree \\), determine the value of x.

\\( x = \square \\)

Explanation:

Step1: Use the property of right - angled triangle

In right - angled triangle \(BFG\), \(\angle BFG = 90^{\circ}\). By the angle - sum property of a triangle (\(\angle BFG+\angle FBG+\angle BGF = 180^{\circ}\)), substituting \(\angle BFG = 90^{\circ}\), we get \(\angle FBG+\angle BGF=90^{\circ}\).

Step2: Substitute the given angle expressions

Since \(m\angle BGF=(7x - 4)^{\circ}\) and \(m\angle FBG=(5x - 14)^{\circ}\), we have the equation \((7x - 4)+(5x - 14)=90\).

Step3: Simplify the left - hand side of the equation

Combine like terms: \(7x+5x-4 - 14=90\), which simplifies to \(12x-18 = 90\).

Step4: Solve for \(x\)

Add \(18\) to both sides of the equation: \(12x-18 + 18=90 + 18\), so \(12x=108\). Then divide both sides by \(12\): \(x=\frac{108}{12}\).

Answer:

\(x = 9\)