Find all angles, osO<360, that satisfy the equation below, to the nearest 10th of a
degree

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

33/100 is what i got

it is definatly the 3rd one. very much looks like it, for a good chance, the third one would be best for you.

8(4)^8=72^feet to say that the total can conclude to 8(4) which as you know would mean 8 x 4. that equals 32w. 32^ could make a 72 answer as well. but your coefficent will remain w so your answer is 72w.

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we know that

major arc is defined as the bigger arc connecting two points on the edge of a circle. measure of major arc is always more than ° or π radian

in this problem

the mayor arc is bac

find the ratio of the measure of the major arc to the measure of the minor arc

therefore

the answer is the option