Question

if O is the centre of a circle of radius r and AB is a chord of circle at distance r/2 from O then angle BAO =?​

Answer 1

Answer:

Since AO = r (radius of circle)

AM =

r2 (given)

Extended OM to D where MD =

r2

Consider the triangles AOM and triangle AMD

OM = MD

∠AMO=∠AMD = 90°

AM = AM (common Sides

So by SSS property

Δ AMO ≅ Δ DM

So AD = AO = r and OD=OM+MD=r

Hence ΔAOD is equilateral triangle

So ∠OAD = 60°

We know that in equilateral triangle altitudes divide the vertex angles

Therefore

= 30°