Question

In the figure given alongside, AB is a diameter of the circle with centre O and CD||BA. If
ZCAB = 400, find,
1.DOC
2.DAC
3.ADC​

Answer 1

Step-by-step explanation:

(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 = 1212∠DOC = 1212 (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 ∆)Read more on Sarthaks.com - https://www.sarthaks.com/976954/the-figure-given-diameter-of-the-circle-with-centre-and-cd-ba-if-cab-find-the-value-cob-doc