Question

In the Figure, O is the Centre of the Circle.
If ∠OAB = 40°, then ∠ACB is equal to

(a) 50°
(b) 40°
(c) 60°
(d) 70°​

Answer 1

$\Huge\underline{\overline{\mid{\green{Answer}}\mid}}$

☞OA=OB (radius of circle)

∠OAB=∠OBA (angles opposite to equal sites)

∠OBA=40°

☞In right angled ΔOAB

∠OAB+∠OBA+∠AOB=180°

40°+40°∠AOB=180°

∠AOB=180°-80°

$∠AOB=100°$

We know that

$∠ACB= \frac{1}{2} ∠AOB$

$= \frac{1}{2} \times 100° = 50°$

☞Thus, (A) is correct.

@gurmanpreet1023

$\Huge \boxed{ \colorbox{gold}{hope \: this \: helps}}$

Answer 2

Given :

• O is the centre of a circle.
• ∠OAB = 40°

To find :

• ∠ACB = ?

Solution :

OA = OB (radii of the same circle)

∠OBA = ∠OAB (Angles opposite to equal sides are equal.)

∠OBA = ∠OAB = 40°

In △AOB

∠OAB + ∠OBA + ∠AOB = 180° (∵Sum of angles of triangle = 180°)

⇒ 40° + 40° + ∠AOB = 180°

⇒ 80° + ∠AOB = 180°

⇒ ∠AOB = 180° - 80°

⇒∠AOB = 100°

Angle at the centre is twice the angle at the circumference.

⇒ 2∠ACB = ∠AOB

⇒ ∠ACB = ∠AOB/2

⇒∠ACB = 100°/2

⇒ ∠ACB = 50°

Answer ⇒ ∠ACB is equal to (a) 50°