QUESTION IMAGE
Question
measure of \\( \angle a b g \\) is always equal to the measure of \\( \angle f b e \\).
sum of the measures of \\( \angle g b f, \angle f b e \\), and \\( \angle d b c \\) is always equal to 1
the sum of the measures of \\( \angle a b g, \angle f b e \\), and \\( \angle d b c \\) is always equal to 180
the measure of \\( \angle g b f \\) is always equal to the measure of \\( \angle d b c \\).
the sum of the measures of \\( \angle a b g, \angle g b f, \angle f b e \\) and \\( \angle d b c \\) is always equal
part b
solve for \\( x \\) given that \\( m \angle a b g = ( 2 x + 3 ) ^ { \circ }, m \angle g b f = ( 5 x + 9 ) ^ { \circ } \\) and \\( m \angle f b e = 42 ^ { \circ } \\).
\\( x = \\)
Step1: Use the angle relationship
Since \(m\angle ABG=m\angle FBE = 42^{\circ}\) (given in the problem statement that the measure of \(\angle ABG\) is always equal to the measure of \(\angle FBE\)), and assume that \(\angle ABG+\angle GBF+\angle FBE = 180^{\circ}\) (a straight - line angle).
Substitute the expressions for the angles: \((2x + 3)+(5x + 9)+42=180\).
Step2: Simplify the left - hand side of the equation
Combine like terms: \(2x+5x+3 + 9+42=180\), which simplifies to \(7x+54 = 180\).
Step3: Solve for \(x\)
Subtract 54 from both sides: \(7x=180 - 54\), so \(7x=126\).
Divide both sides by 7: \(x=\frac{126}{7}\).
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\(x = 18\)