Question

In the figure, is the centre of the circle. If angle AOB - 140° and angle OAC -50°.
Find : () angle ACB (ii) angle OBC () angle OAB (iv) angle CBA​

Answer 1

Answer:

draw a line between A and B then OAB becomes an isosceles triangle with OA=OB (radius of circle)

in triangle AOB

angle O + angle OAB + angle OBA= 180°

as the triangle is isosceles

angle OAB= angle OBA and angle O= 140°

140°+ 2( angle OAB)= 180°

(iii) Angle OAB = 20° and angle OBA= 20°

given angle A = 50°

angle A = angle OAB+ angle BAC

50° = 20°+ angle BAC

angle BAC= 30°

now draw a line from O to C forming an isosceles triangle OAC such that OA= OC

given angle OAC = 50°

that means angle OCA=50°

in triangle OAC sum of angles = 180°

50°+50°+angle AOC =180°

angle AOC= 80°

angle O = angle AOC+ angle BOC

140°= 80°+ angle BOC

angle BOC= 60°

In triangle BOC OC=OB( radius of circle) that means angle OCB = angle OBC

angle BOC+ angle OCB+ angle OBC= 180°

60°+ 2( angle OCB)= 180

angle OCB = 60° (ii) angle OBC= 60°

angle ACB = angle OCA + angle OCB

angle ACB= 50°+ 60° = 110°

(I)angle ACB= 110°

sum of angles in quadrilateral = 360°

angle O + angle A+ angle c + angle B= 360°

140°+50°+110°+angle B= 360°

angle B= 60°

angle B = angle OBA+ angle CBA

60°= 20°+ angle CBA

(iv) angle CBA= 40°