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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 
-7.08 as a fraction in simplest form would be -177/25
We’re told that QR and RS are equal, so this is an isosceles triangle.
Meaning the two base angles must also be equal
Using this information, we know that the third angle must be 180 - (22 + 22), since the angles in a triangle add up to 180
Answer:
-29+7i
Step-by-step explanation:
(5+2i)(-3+i)
-15+5i-6i+2i²
i²=-1
Simplify
-17-i
Add c
c=4i(2+3i)
=8i+12i²
=8i-12
-17-i+8i-12
-29+7i