Question

In the given figure, O is the centre of the circle. If angle DBN = 50 degrees and angle DCM = 60 degrees, find angle BDC. (pls prove without alternate segment theorem as its not part of cbse)


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Answer 1

∠ BDC = 70°

Step-by-step explanation:

We have ∠ OBN = 90° and ∠ DBN = 50° (Given)

So, ∠ DBO = ∠ OBN - ∠ DBN = 90° - 50° = 40° ............ (1)

Again, we have ∠ OCM = 90° and ∠ DCM = 60°( Given)

So, ∠ DCO = ∠ OCM - ∠ DCM = 90° - 60° = 30° .............. (2)

Now, join points D and O.

Then, in Δ DBO, ∠ OBD = ∠ ODB = 40° {from equation (1)}

{Since OD = OB = radius}

Again, in Δ DCO, ∠ OCD = ∠ ODC = 30° {From equation (2)}

{Since, OC = OD = radius}

Therefore, ∠ BDC = ∠ ODB + ∠ ODC = 40° + 30° = 70° (Answer)