In the diagram of circle o, chords `bar(ac)` and `bar(bd)` are perpendicular to each other. if m`/_cod` = 95° what is m`/_aob`?

Given is the circle O, with four points on its circumference A, B, C, and D.

Chords AC and BD are perpendicular to each other at point X. i.e. ∡X is 90° and we have a Right Triangle ΔCXD.

We know that a Central Angle is twice the measure of angle subtended by same arc on its circumference.

It means ∡BOC = 2·∡BDC and ∡DOA = 2·∡ACD

Or we can say ∡BOC = 2·∡XDC and ∡DOA = 2·∡XCD ...............(equation 1)

In Right Triangle ΔCXD, ∡X = 90° ⇔ ∡XCD + ∡XDC = 90°

⇔ 2·∡XCD + 2·∡XDC = 2×90°

Using equation 1;

⇔ ∡BOC + ∡DOA = 180° ......................(equation 2)

We have four Central Angles:- ∡AOB, ∡BOC, ∡COD, and ∡DOA such that:-

∡AOB + ∡COD + ∡BOC + ∡DOA = 360°

Using equation 2;

∡AOB + ∡COD + 180° = 360°

∡AOB + ∡COD = 360° - 180° = 180°

Given ∡COD = 95°

So ∡AOB + 95° = 180°

∡AOB = 180° - 95° = 85°

Hence, ∡AOB = 85° is the final answer.

85 degrees is %100 correct on PLATO

Step-by-step explanation:

< 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⭐

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