Question

in the alongside figure,ABCD is a parallelogram in which DAO =35°,BAO=40° and COD=65°. FIND (a)ABO (b) ODC (c) ACB (d) CBD NOW FIND IT.​

Answer 1

Answer:

Step-by-step explanation:

(i) From the figure we know that ∠ AOB and ∠ COD are vertically opposite angles So we get ∠ AOB = ∠ COD = 105o Consider △ AOB By sum property of a triangle ∠ OAB + ∠ AOB + ∠ ABO = 180o By substituting the values in above equation 35o + 105o + ∠ ABO = 180o On further calculation ∠ ABO = 180o – 35o – 105o By subtraction ∠ ABO = 180o – 140o ∠ ABO = 40o (ii) We know that AB || DC and BD is a transversal From the figure we know that ∠ ABD and ∠ CDB are alternate angles It can be written as ∠ CDO = ∠ CDB = ∠ ABD = ∠ ABO = 40o So we get ∠ ODC = 40o (iii) We know that AB || CD and AC is a transversal From the figure we know that ∠ ACB and ∠ DAC are alternate opposite angles So we get ∠ ACB = ∠ DAC = 40o (iv) We know that ∠ B can be written as ∠ B = ∠ CBD + ∠ ABO So we get ∠ CBD = ∠ B – ∠ ABO In a parallelogram we know that the sum of all the angles is 360o So we get ∠ A + ∠ B + ∠ C + ∠ D = 360o It can be written as 2 ∠ A + 2 ∠ B = 360o By substituting values in the above equation 2 (40o + 35o) + 2 ∠ B = 360o On further calculation 2 (75o) + 2 ∠ B = 360o So we get 150o + 2 ∠ B = 360o 2 ∠ B = 360o – 150o By subtraction 2 ∠ B = 210o By division ∠ B = 105o So we get ∠ CBD = ∠ B – ∠ ABO By substituting values ∠ CBD = 105o – 40o By subtraction ∠ CBD = 65o