Question

7 if j is the incenter of △def, ( mangle dej=(6x + 1) ) and ( mangle jef=(10x - 23) ), find ( mangle fdj ).

Explanation:

Step1: Use the property of in - center (angle bisector)

Since \(J\) is the in - center of \(\triangle DEF\), \(EJ\) is the angle bisector of \(\angle DEF\). So \(m\angle DEJ=m\angle JEF\).

$$6x + 1=10x-23$$

Step2: Solve the equation for \(x\)

Subtract \(6x\) from both sides:

$$1 = 4x-23$$

Add \(23\) to both sides:

$$24=4x$$

Divide both sides by \(4\):

$$x = 6$$

Step3: Find \(m\angle DEF\)

\(m\angle DEJ=(6x + 1)\), substitute \(x = 6\)
\(m\angle DEJ=6\times6+1=37^{\circ}\)
\(m\angle DEF=m\angle DEJ + m\angle JEF=2\times37^{\circ}=74^{\circ}\)

Step4: Use the angle sum property of a triangle

In \(\triangle DEF\), we know that \(\angle EFD = 25^{\circ}\) (given from the figure, assume the non - marked angle at \(F\) part is \(25^{\circ}\)), and \(\angle DEF = 74^{\circ}\)
By the angle sum property of a triangle \(\angle DEF+\angle EFD+\angle FDE=180^{\circ}\)
\(\angle FDE=180-(74 + 25)=81^{\circ}\)

Step5: Use the in - center property (angle bisector) for \(\angle FDE\)

Since \(DJ\) is the angle bisector of \(\angle FDE\) (because \(J\) is the in - center)
\(m\angle FDJ=\frac{1}{2}m\angle FDE\)
\(m\angle FDJ = 23^{\circ}\) (assuming there was a miscalculation in the previous step of angle sum, if we re - check:
If \(x = 6\), \(m\angle DEJ=6\times6 + 1=37^{\circ}\), \(m\angle JEF=10\times6-23 = 37^{\circ}\), \(\angle DEF=74^{\circ}\). If \(\angle EFD = 25^{\circ}\), then \(\angle FDE=180-(74 + 25)=81^{\circ}\) is wrong. Wait, no, looking at the problem again, maybe the \(25^{\circ}\) is \(\angle EFH\) (the non - bisected part). Wait, no, \(J\) is the in - center.
Wait, correct approach:
Since \(J\) is the in - center, \(EJ\) bisects \(\angle DEF\), \(FJ\) bisects \(\angle EFD\), \(DJ\) bisects \(\angle FDE\)
We have \(m\angle DEJ=m\angle JEF\), \(6x + 1=10x-23\), \(x = 6\)
\(m\angle DEJ=37^{\circ}\), \(m\angle DEF = 74^{\circ}\)
If \(\angle EFH = 25^{\circ}\), then \(m\angle EFD=2\times25^{\circ}=50^{\circ}\) (because \(FJ\) is the angle bisector)
By angle sum of triangle \(\angle FDE=180-(74 + 50)=56^{\circ}\)
Since \(DJ\) bisects \(\angle FDE\), \(m\angle FDJ=\frac{1}{2}\times56^{\circ}=28^{\circ}\)

Answer:

\(28^{\circ}\)