Question:
In the adjoining figure, ABCD is a parallelogram in which ∠BAO = 35°, ∠DAO = 40° and ∠COD = 105°. Calculate (i) ∠ABO, (ii) ∠ODC, (iii) ∠ACB, (iv) ∠CBD.
Solution:
ABCD is a parallelogram.
∴ AB ∣∣ DC and BC ∣∣ AD
(i) In ∆AOB, ∠BAO = 35°, ∠AOB = ∠COD = 105° (Vertically opposite angels)
∴ ∠ABO = 180o − (35o + 105o) = 40o
(ii)∠ODC and ∠ABO are alternate interior angles.
∴ ∠ODC = ∠ABO = 40o
(iii) ∠ACB = ∠CAD = 40o (Alternate interior angles)
(iv) ∠CBD = ∠ABC − ∠ABD ...(i)
∠ABC = 180o − ∠BAD (Adjacent angles are supplementary)
⇒∠ABC = 180o − 75o = 105o
⇒∠CBD = 105o − ∠ABD (∠ABD = ∠ABO)
⇒∠CBD = 105o − 40o = 65o
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