In the diagram of circle A, what is m? 75° 90° 120° 135° Posted on October 22, 2021 By Dwighthibbert56 4 Comments on In the diagram of circle A, what is m? 75° 90° 120° 135° In the diagram of circle A, what is m? 75° 90° 120° 135° SAT
[tex]\angle m=90^o[/tex]Step-by-step explanation:The picture of the question in the attached figurewe know thatThe measurement of the external angle is the half-difference of the arches that compriseso[tex]\angle m=\frac{1}{2}(major\ arc\ LN-minor\ arc\ LN)[/tex]Remember that the sum of the major arc and a minor arc is equal to 360 degreeswe have[tex]major\ arc\ LN=270^o[/tex] ----> is givenso[tex]minor\ arc\ LN=360^o-270^o=90^o[/tex]substitute the values[tex]\angle m=\frac{1}{2}(270^o-90^o)=90^o[/tex][tex]Circle A is shown. Tangents L M and N M intersect at point M outside of the circle. Arc L N is 270 d[/tex]Reply
90Step-by-step explanation:Alright first step what is the degree measure of LN.A full rotation is 360 degrees so LN=360-270=90.So angle M is half the difference of the intercepted arcs:[tex]\frac{1}{2}(270-90)=\frac{1}{2}(180)=90[/tex]Reply
[tex]\angle m=90^o[/tex]
Step-by-step explanation:
The picture of the question in the attached figure
we know that
The measurement of the external angle is the half-difference of the arches that comprise
so
[tex]\angle m=\frac{1}{2}(major\ arc\ LN-minor\ arc\ LN)[/tex]
Remember that the sum of the major arc and a minor arc is equal to 360 degrees
we have
[tex]major\ arc\ LN=270^o[/tex] ----> is given
so
[tex]minor\ arc\ LN=360^o-270^o=90^o[/tex]
substitute the values
[tex]\angle m=\frac{1}{2}(270^o-90^o)=90^o[/tex]
[tex]Circle A is shown. Tangents L M and N M intersect at point M outside of the circle. Arc L N is 270 d[/tex]
90
Step-by-step explanation:
Alright first step what is the degree measure of LN.
A full rotation is 360 degrees so LN=360-270=90.
So angle M is half the difference of the intercepted arcs:
[tex]\frac{1}{2}(270-90)=\frac{1}{2}(180)=90[/tex]
B
Step-by-step explanation:
90
It is in the center