Question

Given , a circle with a centre O.BD=OD abd CD perpendicular to AB . Find angle CAB and angle BCD

Answer 1

Question:

Given , a circle with a centre O. BD = OD and CD perpendicular to AB . Find angle CAB and angle BCD.

*See attachment for the original drawing.

Solution:

ΔOBD is an equilateral triangle

Proof:

Given that OB = BD

and OB = OD (both are radius)

⇒ OB = BD = OD  

⇒ ΔOBD is an equilateral triangle

Find ∠AOD:

Angles on a straight line is 180º

∠AOD = 180 - 60

∠AOD =  120º

Find ∠ACD:

Angle at the centre is twice the angle at the circumference

∠ACD = 1/2 ∠AOD

∠ACD = 1/2 (120)

∠ACD = 60º

Find ∠AEC:

Given CD ⊥AB

∠AEC = 90º 

Find ∠CAB:

Sum of angles in the Δ ACD = 180º

∠CAB = 180 - 90 - 60

∠CAB = 30º

Find ∠ACB:

Angle of triangle at the semicircle = 90º

∠CAB = 90º

Find ∠CBA:

Sum of angles in the Δ ACB = 180º

∠CBA = 189 - 90 - 30

∠CBA = 60º

Find ∠CBD:

∠CBD = ∠CBA + ∠ABD

∠CBD = 60 + 60

∠CBD = 120º

Find ∠BCD:

Δ  BCD is an isosceles triangle

∠BCD = (180 - 120) ÷ 2  

∠BCD = 30º

Answer: (A) ∠CAB = 30º (B) ∠BCD = 30º