In the figure given, AB is the diameter of the circle with center O an

1.	In the figure given, AB is the diameter of the circle with center O and CD \|\| BA. If ∠CAB = x, find the value of (i) ∠COB (ii) ∠DOC (iii) ∠DAC (iv) ∠ADC.
Answer» (i) ∠COB = 2 ∠CAB = 2x° (angle at the centre = 2 × angle at the remaining part of the circumference) (ii) ∠OCD = ∠COB = 2x° (alternate ∠s, DC \|\| AB) OD = OC (radii of the same circle) ⇒ ∠OCD = ∠ODC ⇒ ∠ODC = 2x° ∴ In ∆DOC, ∠DOC = 180° – (2x° + 2x°) = 180° – 4x° (∠sum prop. of a ∆) (iii) ∠DAC = \(\frac{1}{2}\)∠DOC = \(\frac{1}{2}\) (180 − 4x)° (angle made by arc DC at the centre = Twice the angle at the remaining part of the circumference) = (90 – 2x)° (iv) In ∆ADC, ∠ACD = ∠CAB = x° (alt ∠s; DC \|\| AB) ∴ ∠ADC = 180° – (x° + 90° – 2x°) = (90 + x)°. (∠sum prop. of a ∆)

In the figure given, AB is the diameter of the circle with center O and CD || BA. If ∠CAB = x, find the value of (i) ∠COB (ii) ∠DOC (iii) ∠DAC (iv) ∠ADC.

Answer»

(i) ∠COB = 2 ∠CAB = 2x° (angle at the centre = 2 × angle at the remaining part of the circumference)

(ii) ∠OCD = ∠COB = 2x° (alternate ∠s, DC || AB)

OD = OC (radii of the same circle)

⇒ ∠OCD = ∠ODC

⇒ ∠ODC = 2x°

∴ In ∆DOC, ∠DOC = 180° – (2x° + 2x°)

= 180° – 4x° (∠sum prop. of a ∆)

(iii) ∠DAC = \(\frac{1}{2}\)∠DOC = \(\frac{1}{2}\) (180 − 4x)°

(angle made by arc DC at the centre = Twice the angle at the remaining part of the circumference)

= (90 – 2x)°

(iv) In ∆ADC, ∠ACD = ∠CAB = x° (alt ∠s; DC || AB)

∴ ∠ADC = 180° – (x° + 90° – 2x°)

= (90 + x)°. (∠sum prop. of a ∆)

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