Question

If O is the centre of a circle of radius r and AB is chord of the circle at a distance r/2 from O, then ∠BAO =
A. 60°
B. 45°
C. 30°
D. 15°

Answer 1

Answer:

The measure of angle ∠BAO is 45°

Step-by-step explanation:

Given as :

For circle with center O

The radius of circle = r

AB is chord at the distance of $\dfrac{r}{2}$  from center O

With chord AB and center O , Δ BOA is formed

for Δ BOA , OA = OB = r   ( radius of same circle )

And straight line of line $\dfrac{r}{2}$ bisect the chord

∠ BOA  = 90°

OA = OB = r ( radius of same circle )

As the sum of three angles of any triangle = 180°

i.e ∠BOA + ∠BAO + ∠ABO = 180°

And  ∠ BOA  = 90°

And OA = OB , so the Triangle must be isosceles

For isosceles Triangle , two sides are common and opposite angle must be same .

i.e  ∠BAO = ∠ABO

And from the property of isosceles Triangle, ∠BAO + ∠ABO = 180° - 90°

i.e   ∠BAO + ∠ABO =  90°

and ∠BAO = ∠ABO

So, ∠BAO = ∠ABO = 45°

Hence, The measure of angle ∠BAO is 45° . Answer

Answer 2

The value of ∠BAO = 30°

Step-by-step explanation:

Let the radius of the circle be r

From diagram, OP ⊥ AB

From ΔOPA,

OP = r/2 and OA = r

OP = 1/2 OA

In 30°-60°-90° triangle, if opposite side angle is 30° then opposite side angle is half of hypotenuse.

Side opposite to ∠OAP = 1/2 Hypotenuse

∠OAP = 30°

∴ ∠OAB = 30°