Answer:
The measure of [tex]m \hat {BC}=34^0[/tex]
The measure of [tex]\angle DFE = 81^0[/tex]
Step-by-step explanation:
The correct question is added in the diagram below.
From the diagram;
Given that:
∠A = 55°
∠P =38°
[tex]m \hat {AC} = 86^0[/tex]
[tex]m \hat{DE} = 2( \angle A )[/tex]
[tex]m \hat{DE} = 2( 55^0 )[/tex]
[tex]m \hat{DE} = 110^0[/tex]
a) it is clear and obvious that ∠P is formed at the exterior of the circle; so:
[tex]m \angle P = \dfrac{m \hat {DE} -m \hat {BC} }{2}[/tex]
[tex]38= \dfrac{110 -m \hat {BC} }{2}[/tex]
[tex]38*2={110 -m \hat {BC} }{[/tex]
[tex]76={110 -m \hat {BC} }{[/tex]
[tex]m \hat {BC} }{={110 -76[/tex]
[tex]m \hat {BC}=34^0[/tex]
b) [tex]m \hat {AC}=m \hat {AB}+ m \hat {BC}[/tex]
[tex]86=m \hat {AB}+ 34[/tex]
[tex]m \hat {AB}=86-34[/tex]
[tex]m \hat {AB}=52[/tex]
∴ [tex]\angle DFE = \dfrac{m \hat{DE}+m \hat{AB}}{2}[/tex] (rule: chord intersecting inside the circle)
[tex]\angle DFE = \dfrac{110+52}{2}[/tex]
[tex]\angle DFE = \dfrac{162}{2}[/tex]
[tex]\angle DFE = 81^0[/tex]
