Question

In the figure, O is the centre of the circle. If angle ADC = 140°, then what is the value of x?​

Answer 1

Given:

◉ O is the center of the circle

angle ADC=140 degree

To find :

➡ the value of x=?

Solution :

In the following figure shows

➡ OA=OC

$\\ \implies\sf{\angle OAC= \angle ACO =x}$

$\\ \implies\sf{\angle ADC+ \angle AMC =180 \degree ....(cyclic \: quadrilateral) }$

$\\ \implies\sf{ \angle AMC =180 \degree - 140 \degree =40 \degree}$

$\\ \implies\sf{ \angle ADC =2 \angle AMC =2 \times 40 =80 \degree}$

In triangle ADC,

$\\ \implies\sf{ \angle ADC + \angle OAC +\angle OCA =180 \degree}$

$\\ \implies \sf \: x + 80 \degree + x = 180 \degree \\ \\ \implies \sf \: 2x + 80 \degree = 180 \degree \\ \\ \implies \sf \: 2x + 80 \degree - 180 \degree = 0 \\ \\ \implies \sf \: 2x - 100 \degree = 0 \\ \\ \implies \sf \: 2x = 100 \degree \\ \\ \implies \sf \: x = \frac{100 \degree}{2} = 50 \degree \\ \\ \implies \boxed{ \sf{x = 50 \degree}}$