consider the quadrilateral formed by the vertices a,b,c and o the centre of the circle

aoc = 40 degrees

at the suggest of call tangents space perpendicular to radius therefore other 2 angles are 90 each.

You are watching: If the measure of arc ac is 160°, what is the measure of angle abc?

hence the 4th angle abc = 360-(90+90+40) =140

(since amount of angle of square is 360)

m∠ABC=20°

Step-by-step explanation:

Consider square ABCO. The sum of the actions of all inner angles in square ABCO is equal to 360°.

Lines BA and BC space tangent to the circle, climate the radii OC and also OA space perpendicular to the tangent present BC and BA. Therefore,

m∠BCO=90°;m∠BAO=90°.

The measure up of the edge AOC is 160° (because the measure of arc AC is 160°). So,

m∠ABC+m∠BCO+m∠BAO+m∠AOC=360°,

m∠ABC=360°-(m∠BCO+m∠BAO+m∠AOC)=360°-(90°+90°+160°)=20°. ∠ABC = 80°

A is correct.

Step-by-step explanation:

If the measure up of arc AC is 160°.

Angle subtended on one is half the edge subtended at facility on exact same arc.

Please express the attachment because that figure.

Therefore, where, arc AC = 160°  Hence, the measure of ∠ABC is 80°  Tangent lines of a circle space lines that just pass one allude on the circle. Due to the fact that you forgot to include the picture, I portrayed my own as what the problem described it to be.One that the theorems of circle is that the measure up of the intercepted arc is double as that of its the contrary angle. In this case, because the intercepted arc steps 160°, then the measure up of angle alphabet is 160/2 = 80°. Therefore, the price is 80°. The angle formed by the intersection the 2 tangents outside the circle equals fifty percent the distinction of the intercepted arcs.arc AC = 160° (given)arc ADC = 360 - 160 = 200°∠ABC = (200-160)/2= 20° ← answer The answer to the trouble is d.20

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