А
30°
23. In the adjoining figure, O is the centre of the circle. Tangents
to the circle at A and B meet at C. If /_ACO = 30°, find
(1) /_BCO
(ii) /_AOB
(iii) /_APB.
Answers
Answer:
In circle with centre O, CA and CB are tangent segments.
OA =~ OB - (i) (radii of same circle)
seg CA and seg CB are tangents are A and B resp.
:. OA is perpendicular to CA - Tangent Theorem
:. OB is perpendicular to CB - Tangent Theorem
:. /_OAC and /^ OBC = 90°
Also, seg CA =~ seg CB. -(ii) Tangent Segment Theorem
Now, In triangle ACO and BCO,
seg CA =~ seg CB - from (ii)
seg OA =~ seg OB - from (i)
seg OC =~ seg OC - common side
:. triangle ACO =~ BCO - (SSS test of congruency)
:. /_ ACO =~ /_ BCO - (c.a.c.t.)
:. /_ BCO = 30°
Now, in triangle ACO,
/_ACO + /_ AOC + /_ OAC = 180° - sum of measures of all angles in a triangle.
30° + 90° + /_ AOC = 180°
:. /_ AOC = 60°
Similarly, /_BOC can also be proved as 60°
:. /_ AOB = /_AOC + /_ BOC
:. /_ AOB = 120°
/_AOB is the central angle .
:. m(arc AB) = 120° - Definition of Minor arc
/_ APB is intercepted in arc AB
:. /_APB = 1/2 m(arc AB) - (I don't remember exact reason)
:. /_APB= 60°
Hence, all the answers are being found out.
I hope it helps.
Mark this as the Brainliest.