Answer:
Problem 1)
[tex]m\angle DEG = 38^\circ[/tex]
Problem 2)
[tex]m\angle x =140^\circ \text{ and } m\angle y = 49^\circ[/tex]
Problem 3)
[tex]m\angle EDG = 65^\circ[/tex]
Step-by-step explanation:
Problem 1: ∠DEG
From the diagram, we know that ∠DFG intercepts Arc DG.
Inscribed angles have half the measure of its intercepted arc. Therefore:
[tex]\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=m\angle DFG[/tex]
We know that m∠DFG = 38°. So:
[tex]\displaystyle \frac{1}{2}m\stackrel{\frown}{DG}=38\Rightarrow m\stackrel{\frown}{DG}=76^\circ[/tex]
∠DEG also intercepts Arc DG. Hence:
[tex]\displaystyle m\angle DEG=\frac{1}{2}m\stackrel{\frown}{DG}[/tex]
We know that Arc DG measures 76°. Hence:
[tex]\displaystyle m\angle DEG =\frac{1}{2}\left(76\right)=38^\circ[/tex]
Alternate Explanation:
Since ∠DEG and ∠DFG intercept the same arc, ∠DEG ≅ DFG. So, m∠DEG = m∠DFG = 38°.
Problem 2: Circle with Centre O
(Let the bottom left corner be A, upper be B, and right be C.)
∠ACB intercepts Arc AC.
Inscribed angles have half the measure of its intercepted arc. Therefore:
[tex]\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=m \angle ACB[/tex]
Since m∠ACB = 70°:
[tex]\displaystyle \frac{1}{2}m\stackrel{\frown}{AC}=70\Rightarrow m\stackrel{\frown}{AC}=140^\circ[/tex]
∠x is a central angle and also intercepts Arc AC.
The measure of a central angle is equal to its intercepted arc. Thus:
[tex]\displaystyle m\angle x =m\stackrel{\frown}{AC}=140^\circ[/tex]
The sum of the interior angles of a polygon is given by the formula:
[tex](n-2)\cdot 180^\circ[/tex]
Where n is the number of sides.
Since the inscribed figure is a four-sided polygon, its interior angles must total:
[tex](4-2)\cdot 180^\circ =360^\circ[/tex]
Therefore:
[tex]21+70+y+(360-x)=360[/tex]
Substitute and solve for y:
[tex]91+y+(360-140)=360\Rightarrow y +311=360\Rightarrow m\angle y = 49^\circ[/tex]
Problem 3: ∠EDG
∠EFG intercepts Arc EDG.
Inscribed angles have half the measure of its intercepted arc. Therefore:
[tex]\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG}=m\angle EFG[/tex]
Since m∠EFG = 115°:
[tex]\displaystyle \frac{1}{2}m\stackrel{\frown}{EDG} = 115\Rightarrow m\stackrel{\frown}{EDG} = 230^\circ[/tex]
A full circle measures 360°. Hence:
[tex]m\stackrel{\frown}{EDG}+m\stackrel{\frown}{EFG}=360^\circ[/tex]
Since we know that Arc EDG measures 230°:
[tex]230^\circ + m\stackrel{\frown}{EFG}=360[/tex]
Solve for Arc EFG:
[tex]m\stackrel{\frown}{EFG}=130^\circ[/tex]
∠EDG intercepts Arc EFG.
Inscribed angles have half the measure of its intercepted arc. Therefore:
[tex]\displaystyle m\angle EDG = \frac{1}{2}m\stackrel{\frown}{EFG}[/tex]
Since we know that Arc EFG measures 130°:
[tex]\displaystyle m\angle EDG = \frac{1}{2}(130)=65^\circ[/tex]
