Question

 In the figure, O is the centre of the circle. What is the value of x?

       (i) 125°

(ii) 105°

(iii) 95°

(iv) 85°

Answer 1

$\huge\underline\mathfrak{Answer:-}$

50°


$\huge\underline\mathfrak{Solution:-}$



$\huge\underline\mathfrak{Theorm 1:-}$

The angle subtended at the centre of a circle by an arc
is double the angle subtended by it on any point on the
remaining part of the.

$\huge\underline{Step 1:-}$

According to First Theorm :-

$\angle{AOB}$=2$\angle{ACB}$

$\textbf{Therefore,}$

$\angle{AOB}$=2(40°)

$\angle{AOB}$=80°




$\huge\underline{Step 2:-}$

Now in ∆AOB

$\huge\underline\mathfrak{Theorm 2:-}$

AO=OB.................
$\textbf{Radius of circle}$

$\textbf{Therefore,}$

AO=x

OB=x





$\huge\underline{Step 3:-}$

In ∆ ACB
$\huge\underline\mathfrak{Theorm 3:-}$

Sum of angle of ∆=180°

$\textbf{Therefore,}$

$\angle{AOB}$+$\angle{OBA}$+

$\angle{BAO}$=180°

On putting all values

=80°+x+x=180°

=2x=100°

$=x = \frac{100}{2}$

X=5O°

Answer 2

Answer:

The answer is below

Step-by-step explanation:

Answer:−

50°

\hugemathfrak{Solution:-}

Solution:−

\huge\italic\mathfrak{Theorm 1:-}

Theorm1:−

The angle subtended at the centre of a circle by an arc

is double the angle subtended by it on any point on the

remaining part of the.

\huge\bold\{Step 1:-}

Step1:−

According to First Theorm :-

\angle{AOB}∠AOB =2\angle{ACB}∠ACB

\textbf{Therefore,}Therefore,

\angle{AOB}∠AOB =2(40°)

\angle{AOB}∠AOB =80°

\huge\underline{Step 2:-}

Step2:−

Now in ∆AOB

\huge\underline\mathfrak{Theorm 2:-}

Theorm2:−

AO=OB.................

\textbf{Radius of circle}Radius of circle

\textbf{Therefore,}Therefore,

AO=x

OB=x

\huge\underline{Step 3:-}

Step3:−

In ∆ ACB

\huge\underline\mathfrak{Theorm 3:-}

Theorm3:−

Sum of angle of ∆=180°

\textbf{Therefore,}Therefore,

\angle{AOB}∠AOB +\angle{OBA}∠OBA +

\angle{BAO}∠BAO =180°

On putting all values

=80°+x+x=180°

=2x=100°

=x = \frac{100}{2}=x=

2

100

X=5O°