Question

1. if ∠cde is a straight angle. de bisects ∠gdh, m∠gde=(8x - 1)°, m∠edh=(6x + 15)°, and m∠cdf = 43°. find each measure.

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{DE}$ 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 $m\angle{GDE}=8x - 1$, substituting $x = 8$ 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, so $m\angle{CDE}=180^{\circ}$, and $m\angle{CDF} = 43^{\circ}$. Also, $m\angle{GDH}=126^{\circ}$. First, find the non - overlapping part of the angles. $m\angle{FDH}=180^{\circ}-m\angle{CDF}-m\angle{GDH}$. Substituting the values, we have $m\angle{FDH}=180^{\circ}-43^{\circ}-126^{\circ}=11^{\circ}$.

Step4: Find $m\angle{FDE}$

$m\angle{FDE}=m\angle{FDH}+m\angle{EDH}$. Since $m\angle{EDH}=6x + 15$ and $x = 8$, $m\angle{EDH}=6\times8+15=63^{\circ}$, and $m\angle{FDH}=11^{\circ}$, then $m\angle{FDE}=11^{\circ}+63^{\circ}=74^{\circ}$.

Answer:

$x = 8$
$m\angle{GDH}=126^{\circ}$
$m\angle{FDH}=11^{\circ}$
$m\angle{FDE}=74^{\circ}$