Question

If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then calculate ∠BAO.
​

Answer 1

Answer:

isosceles Δ BOA,

so, ∠ BAO = ∠ABO,

now, by ASP of a Δ,

∠BAO+∠ABO+∠AOB=180°

⇒2∠BAO=180°-60°

⇒2∠BAO=120°

⇒∠BAO=60°

∴ all angles in ΔBOA are 60°

⇒ΔBOA is equilateral

∴ OA=OB=AB

⇒AB= radius.                              (∵ OA and OB are radii)

thank you

Answer 2

Given :

• O is the centre of a circle.
• Radius of a circle is r.
• Chord of a circle is AB at a distance r/2 from O.

To calculate :

• Calculate the ∠BAO = ?

Solution :

To find the ∠BAO = ?

$\implies \bf OP \ = \ \dfrac {r}{2}$

$\implies \bf AO \ = \ r$

$\implies \bf AP^2 \ = \ r^2 \ - \ \dfrac {r^2}{4}$

$\implies \bf AP \ = \ \dfrac {3r^2}{4}$

$\implies \bf AP \ = \ \dfrac {\sqrt {3}r}{2}$

Here, We have to find ∠BAO = ?

So,

$\implies \bf Cos A \ = \ \dfrac {OP}{AO} \ = \ \dfrac {\dfrac {r}{2}}{r}$

$\implies \bf \dfrac {1}{2}$

$\implies \bf Tan A \ = \ \dfrac {r}{2} \times \dfrac {2}{\sqrt {3}r}$

$\implies \bf Tan A \ = \ \dfrac {1}{\sqrt {3}}$

$\implies \bf Tan A \ = \ Tan \ 30^{\circ}$

$\qquad \large{\red{\underline {\boxed {\bf \angle BAO \ = \ 30^{\circ}}}}}$