The slopes of lines pq and p’q’ can be determined using the formula the product of these slopes is this product shows that the slopes are negative reciprocals. it is given that the lines are perpendicular and we have shown that the slopes of the lines are negative reciprocals.

we know that

If two lines are perpendicular. then the product of their slopes is equal to -1 because the slopes are negative reciprocal

so

m1*m2=-1

m1=-1/m2

therefore

the answer is

The product of these slopes is -1

< 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⭐

so then what are you asking?

step-by-step explanation: