Question

in the figure , O is the centre of the circle find the value of x​

Answer 1

Answer:

140

Step-by-step explanation:

OCA=40 (isosceles triangle)

OCB=30 ("")

ACB = 70

by theorem

AOB=2*70 = 140

Answer 2

$\underline{ \sf \huge \pink☯ \blue {Answer:}}$

140°

$\sf \huge \purple { \underline{Explanation:}}$

In ∆AOC

$\sf \angle \: A = \angle \: ACO \: \: \: \\ \sf(\because equal \: angle \: have \: equal \: opposite \: side) \\ \\ \sf\angle \: ACO = 40 \degree$

Similarly in ∆BOC:-

$\sf \angle \:BCO \: = 30 \degree$

$\sf \angle ACB= \angle ACO+ \angle BCO \\\\ \sf \angle ACB=40°+30° \\\\ \sf\angle ACB=70° \\ \\ \sf\angle AOB=2 \angle ACB(Theorem) \\ \\ \angle \sf AOB = 2(70 \degree) \\ \\ \sf\angle AOB = 140 \degree$

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