Question

in the given figure AB is a diameter of a circle with center O. If ADE and CBE are straight lines meeting at E such that ∠BAD=35° and ∠BED=25°. Find ∠DBC, ∠DCB, ∠BDC.
Please answer with steps because this a 5 mark question.

Answer 1

... BAD = 35° and ∠BED = 25°, find (i) ∠DBC (ii) ∠DCB (iii) ∠BDC ...


AB is a diameter of the circle with centre O.
ADE and CBE are straight lines that meet at E such that ∠BAD = 35° and ∠BED = 25°.
Join BD and AC.

(i)
Now, ∠BDA = 90°           (Angle in the semicircle)
Also, ∠EDB + ∠BDA = 180°   (Linear pair)
Or ∠EDB = 90°  

Now, ∠EBD = {180° – (∠EDB + ∠BED}  (Angle sum property)
                  = (180° – (90° + 25°)
                  = (180° – 115°) = 65°
∠DBC = (180° – ∠EBD) = (180° - 65°) = 115°      (Linear pair)
∴ ∠DBC = 115°

(ii)
Here, ∠DCB = ∠BAD        (Angles in the same segment)
∠BAD = 35°
 ∴ ∠DCB = 35°
(iii)
∠BDC = 180° – (∠DBC + ∠DCB)    (Angle sum property)
            = {180° – (∠DBC + ∠BAD)}
            = 180° – (115° + 35°)
            =180° – 150°
            = 30°
∴ ∠BDC = 30°

Hope this helps you

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