Question

1. if ∠cde is a straight angle, (overline{de}) bisects ∠gdh, (mangle gde=(8x - 1)^{circ}), (mangle edh=(6x + 15)^{circ}), and (mangle cdf = 43^{circ}), find each measure.

Explanation:

Step1: Use angle - bisector property

Since $\overline{DE}$ bisects $\angle{GDH}$, then $m\angle{GDE}=m\angle{EDH}$. So, we set up the equation $8x - 1=6x + 15$.
$8x-6x=15 + 1$
$2x=16$
$x = 8$

Step2: Find $m\angle{GDH}$

Since $m\angle{GDH}=m\angle{GDE}+m\angle{EDH}$ and $m\angle{GDE}=m\angle{EDH}$, and $m\angle{GDE}=8x - 1$. Substitute $x = 8$ into $m\angle{GDE}$, we get $m\angle{GDE}=8\times8-1=63^{\circ}$. Then $m\angle{GDH}=2\times m\angle{GDE}=2\times63^{\circ}=126^{\circ}$

Step3: Find $m\angle{FDH}$

We know that $\angle{CDE}$ is a straight - angle ($m\angle{CDE}=180^{\circ}$) and $m\angle{CDF}=43^{\circ}$. Also, $m\angle{GDH}=126^{\circ}$. First, find the non - overlapping part of $\angle{CDF}$ and $\angle{GDH}$.
The sum of angles around point $D$ is $360^{\circ}$. But we can also use the fact that $m\angle{FDH}=180^{\circ}-m\angle{CDF}-m\angle{EDH}$. Since $x = 8$, $m\angle{EDH}=6x + 15=6\times8+15=63^{\circ}$. So $m\angle{FDH}=180^{\circ}-43^{\circ}-63^{\circ}=74^{\circ}$

Step4: Find $m\angle{FDE}$

$m\angle{FDE}=m\angle{FDH}+m\angle{EDH}$. Substitute $m\angle{FDH}=74^{\circ}$ and $m\angle{EDH}=63^{\circ}$, we get $m\angle{FDE}=74^{\circ}+63^{\circ}=137^{\circ}$

Answer:

$x = 8$
$m\angle{GDH}=126^{\circ}$
$m\angle{FDH}=74^{\circ}$
$m\angle{FDE}=137^{\circ}$