we know that
The measurement of the exterior angle is the semi-difference of the arcs which comprises
In this problem
∠FGH is the exterior angle
∠FGH=
∠FGH=
-----> equation A
--------> equation B
Substitute equation B in equation A
![100\°=(arc\ FEH-[360\°-arc\ FEH])](https://tex.z-dn.net/?f=100%5C%C2%B0%3D%28arc%5C%20FEH-%5B360%5C%C2%B0-arc%5C%20FEH%5D%29)
therefore
<u>The answer is</u>
The measure of arc FEH is equal to 
They are alternate Exteror angles and internet but as for the question they would be corresponding angles
Answer:
Measure of arc AE = 58°
Step-by-step explanation:
As shown: ABCD is a quadrilateral, ∠C = 119°
So, ∠C + ∠A = 180°
∴ ∠A = 180° - ∠C = 180° - 119° = 61°
ΔAGB is a right triangle at G
So, ∠A + ∠B = 90°
∴ ∠ABG = 90 - ∠A = 90 - 61 = 29°
Arc AE opposite to the angle ∠ABG
So, measure of arc AE = 2∠ABG = 2 * 29° = 58°
Answer:
5x +11y +10
Step-by-step explanation:
7x+11y+9-2x+1
Combine like terms
7x-2x + 11y +9 +1
5x +11y +10
