Question

in the figure, \\( \overrightarrow { b a } \\) and \\( \overrightarrow { b c } \\) are opposite rays. \\( \overrightarrow { b h } \\) bisects \\( \angle e b c \\) and \\( \overrightarrow { b e } \\) bisects \\( \angle a b f \\).
if \\( m \angle e b c = ( 3 1 a - 2 ) ^ { \circ } \\) and \\( m \angle e b h = ( 4 a + 4 5 ) ^ { \circ } \\), find \\( m \angle h b c \\).

Explanation:

Step1: Use the angle - bisector property

Since \(\overrightarrow{BH}\) bisects \(\angle EBC\), we have \(m\angle EBC = 2m\angle EBH\).
Given \(m\angle EBC=(31a - 2)^{\circ}\) and \(m\angle EBH=(4a + 45)^{\circ}\), then \((31a-2)=2(4a + 45)\).

Step2: Solve the equation for \(a\)

Expand the right - hand side: \(31a-2 = 8a+90\).
Subtract \(8a\) from both sides: \(31a-8a-2=8a - 8a+90\), so \(23a-2 = 90\).
Add \(2\) to both sides: \(23a-2 + 2=90 + 2\), then \(23a=92\).
Divide both sides by \(23\): \(a=\frac{92}{23}=4\).

Step3: Find \(m\angle EBH\)

Substitute \(a = 4\) into \(m\angle EBH=(4a + 45)^{\circ}\).
\(m\angle EBH=(4\times4 + 45)^{\circ}=(16 + 45)^{\circ}=61^{\circ}\).

Step4: Find \(m\angle HBC\)

Because \(\overrightarrow{BH}\) bisects \(\angle EBC\), \(m\angle HBC=m\angle EBH\).

Answer:

\(61\)