Questions: What is the value of (z) ? (z=)°

Solution Steps

Quadrilateral FIGH is an isosceles trapezoid, so $FI \parallel HG$. The angles $FIH$ and $GHI$ are supplementary, so $m\angle IHG + m\angle FIH = 180^\circ$ $m\angle IHG + 50^\circ = 180^\circ$ $m\angle IHG = 180^\circ - 50^\circ$ $m\angle IHG = 130^\circ$

The sum of the angles in quadrilateral FIGH is $360^\circ$. Since $FI=FH$ and $GH=FG$, it is an isosceles trapezoid, and $m\angle GFI = m\angle HFG=z$ and $m\angle FIH=50^\circ$ and $m\angle FHG=98^\circ$. The sum of the angles in the quadrilateral is $360^\circ$. So, $z+z+50+98 = 360$. $2z+148=360$ $2z = 360 - 148$ $2z = 212$ $z = \frac{212}{2}$ $z=106$

The sum of the angles in triangle FIG is $180^\circ$. So, $m\angle FIG + m\angle IGF + m\angle GFI = 180^\circ$ $50^\circ + m\angle IGF + z = 180^\circ$ $50^\circ + m\angle IGF + 106 = 180^\circ$ $156^\circ + m\angle IGF = 180^\circ$ $m\angle IGF = 180^\circ - 156^\circ$ $m\angle IGF = 24^\circ$.

The sum of the angles in the quadrilateral FIGH is $360^\circ$. $50+z+x+98=360$ $148 + 2z=360$ $2z = 212$ $z=106$

Final Answer