Question

1.in the adjoining figure o is the centre of the circle if <aoc = 150° find (i) <abc (ii) <adc

2. in the adjoining figure abcd is a cyclic quadrilateral find the value of x ​

Answer 1

1. given <aoc = 150

now <abc = 1/2×150 = 75

<adc = 180 - 75 =105

2. 2x + 40 + 4x - 64 = 180

so 6x -24 = 180

so x - 4 = 180/6 = 30

so x = 30 +4 = 34

Answer 2

Q1)

2x + 40⁰ + 4x -64⁰ = 180⁰

6x - 24⁰ = 180⁰

6x = 180⁰ + 24⁰

x = 204⁰/6

x = 34⁰

Q2)

2∠ABC = ∠AOC

2∠ABC = 150⁰

∠ABC = 150⁰/2

∠ABC = 75⁰

Then,

∠ABC + ∠ADC = 180⁰

75⁰ + ∠ADC = 180⁰

∠ADC = 180⁰ - 75⁰

∠ADC = 105⁰