Question

give me a answer of (i) and (iii)

Answer 1

(iii)We know that angle subtended by a triangle in a semicircle is 90°.
Hence, ∠BAD=90°.

InΔABD,
         ∠BAD+∠ADB+∠DBA=180
                    90+60+∠ADB=180
                                ∠ADB=180-150
                                ∠ADB=30°.



(i) O is the centre of the circle.
    ∠OCQ=90°          [∵PQ is a tangent]
    ∴∠OCD+∠DCQ=90
            ∠OCD+40=90
               ∠OCD=50°=∠ODC     [∵ΔODC is isosceles as radius is same]

     2∠OCD+∠DOC=180
                   ∠DOC=80°
       
∠DOC+∠BOC=180      [Linear pair]
            ∠BOC=100°

ΔBOC is isosceles,
∴2∠DBC+∠BOC=180
              2∠DBC=80
                ∠DBC=40°

Hope this will help