Answer:
Step-by-step explanation:
<u><em>The picture of the question in the attached figure</em></u>
we know that
The measure of the external angle is the semi-difference of the arches it covers.
so
![m\angle GET=\frac{1}{2}[arc\ TN-arc\ TG]](https://tex.z-dn.net/?f=m%5Cangle%20GET%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20TN-arc%5C%20TG%5D)
Remember that the diameter divide the circle into two equal parts
In this problem
TN is a diameter
we have
----> because is half the circle (TN is a diameter)
---> is given
substitute
![m\angle GET=\frac{1}{2}[180^o-46^o]=67^o](https://tex.z-dn.net/?f=m%5Cangle%20GET%3D%5Cfrac%7B1%7D%7B2%7D%5B180%5Eo-46%5Eo%5D%3D67%5Eo)
Answer:
because your grandfather has the experience to do work
Answer:
x=44, They are adjacent
Step-by-step explanation:
since this is right angle it is 90 degrees. You set the whole thing equal to 90, so 43+x+3=90, add 43 and 3=46. Then you subtract 90-46, you get x=44.
The answer is H to your question
