Question

6. In the given figure, A, B, C are 3 points on
the circumference of a circle with centre O.
Arc AB = 2 arc BC and ZAOB = 108°
Calculate
(i) ZACB (ii) ZBOC (iii) ZOAB
IC С
108°​

Answer 1

Answer:

Refer Figure:

Given that: ∠OAC=53°

∠CBO=32°

Consider △ACO:

AO=CO (radius of circle)

Thus, ∠ACO=∠OAC=53° (angles opposite to equal sides are equal)

∠AOC=180°−53°−53°=74° (Sum of three angles of a triangle is 180°)

Similarly in △BCO:

BO=OC

Hence, ∠OBC=∠OCB

And ∠COB=180°−32°−32°=116°

Reflex∠AOB=∠AOC+∠COB=74°+116°=190°

Thus ∠AOB=360°−reflex∠AOB

∠AOB=360°−190°=170°

Answer: ∠AOB=170°

solution

Answered By

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