Question

In a circle with centre O, AD is a diameter and AC is a chord. B is a point on AC, such that OB = 5 cm and ∠OBA =60°. If ∠DOC= 60°, then what is the length of BC ?

Answer 1

Answer:

length of BC  = 5 cm

Step-by-step explanation:

ΔDOC

∠DOC= 60°

& OC = OD  ( Radius)

=> ∠OCD = ∠ODC = 60°

=> ΔDOC is Equilateral triangle with side = radius

∠ACD = 90°  (inscribed angle by diameter)

=> ∠OAC = 180° - 90° - 60° = 30°

AC² + CD² = AD²

=> AC² + R² = (2R)²

=> AC = √3 R

∠OAC =  30°  ∠OBA =60°

=> ∠AOB = 90°

Draw CE ⊥ AD

(1/2)*AC * CD = (1/2) * AD * CE

=> √3 R * R = 2R * CE

=> CE = √3 R/2

AE² = AC² - CE² = 3R² - 3R²/4 = 9R²/4 = 3R/2

ΔAOB ≈ ΔAEC

=> AB/AC = AO/AE  = OB/CE

=> AB/√3 R = R/(3R/2)  = 5/(√3 R/2)  = K

=> AB/√3 R = 2/3  = 10/(√3 R)  = K

=> 2/3  = 10/(√3 R)

=> R = 5√3

AB/√3 R = 2/3

=> AB  = 2R/√3

=> AB = 10

AC = √3 R  = 15

BC = AC - AB = 15 - 10 = 5

length of BC  = 5 cm

Another Easy Solution

∠ACD = 90°  (inscribed angle by diameter)

=> ∠OCD = 60°  (equilateral triangle)

=> ∠ACO = ∠BCO = 90° - 60° = 30°

∠OBA = ∠BOC + ∠BCO

=> 60° = ∠BOC + 30°

=> ∠BOC  = 30°

in ΔOBC

∠BOC = ∠BCO = 30°

=> OB = BC

=> BC = 5 cm