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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 = 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
Answer:
(i) Bobby waited in front of the school.
(ii) The children were told to return home before it became dark.
(iii) He wore an old muffler wrapped around his
neck.
(iv) We saw the bird fly high above the tower and disappear.
(v) Music therapy is a cure to many ills.
(vi) The thief ran to the staircase and out into the yard.
(vii) Mr. Roy jumped into his car and drove it as fast as he
could.
(viii) Please do not get carried away by these rumours.