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.

6x + 15+43 = 8x - 1
6x + 58 = 8x - 1
+ 1
+ 1
x = 29
6x + 59 = 8x

• 6
• 6

59 = 2x
29
m∠gdh=
____
m∠fdh=
____
m∠fde=
____

Explanation:

Step1: Use angle - bisector property

Since \(DE\) bisects \(\angle GDH\), then \(m\angle GDE=m\angle EDH\). So we set up the equation \(8x - 1=6x + 15\).
\[

$$\begin{align*} 8x-1&=6x + 15\\ 8x-6x&=15 + 1\\ 2x&=16\\ x&=8 \end{align*}$$

\]

Step2: Find \(m\angle GDH\)

Since \(m\angle GDH=m\angle GDE + m\angle EDH\) and \(m\angle GDE=m\angle EDH = 6x+15\) (or \(8x - 1\)), substituting \(x = 8\) into \(m\angle GDE\) (or \(m\angle EDH\)), we get \(m\angle GDE=6\times8 + 15=48+15 = 63^{\circ}\), so \(m\angle GDH=2\times63^{\circ}=126^{\circ}\).

Step3: Find \(m\angle FDH\)

We know that \(\angle CDE\) is a straight - angle (\(180^{\circ}\)) and \(m\angle CDF = 43^{\circ}\), \(m\angle EDH=63^{\circ}\). Then \(m\angle FDH=180^{\circ}-m\angle CDF - m\angle EDH=180 - 43-63=74^{\circ}\).

Step4: Find \(m\angle FDE\)

\(m\angle FDE=m\angle FDH + m\angle EDH=74^{\circ}+63^{\circ}=137^{\circ}\).

Answer:

\(m\angle GDH = 126^{\circ}\), \(m\angle FDH=74^{\circ}\), \(m\angle FDE = 137^{\circ}\)