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ABCD is a parallelogram in which angle $\angle A = {70^0}$ . Compute angle B,angle C and angle D.





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Hint: Use geometrical theorems related to parallel lines.
Given in the problem ABCD is a parallelogram with angle $\angle A = {70^0}$.
We know that opposite sides of the parallelogram are parallel to each other.
$
   \Rightarrow AB{\text{ is parallel to }}CD{\text{ (1)}} \\
   \Rightarrow AD{\text{ is parallel to B}}C{\text{ (2)}} \\
$
Same side Interior angle theorem states that the interior angles formed by a transversal line intersecting two parallel lines are supplementary.
Using the same theorem in parallelogram $ABCD$, we get:
\[
  (1) \Rightarrow {\text{ }}\angle A + \angle D = {180^0}{\text{ and }}\angle B + \angle C = {180^0} \\
  (2) \Rightarrow {\text{ }}\angle A + \angle B = {180^0}{\text{ and }}\angle C + \angle D = {180^0} \\
 \]
Using angle $\angle A = {70^0}$ in above equations, we get:
  \[
  {70^0} + \angle D = {180^0}{\text{ }} \\
   \Rightarrow \angle D = {110^0} \\
  {70^0} + \angle B = {180^0} \\
   \Rightarrow \angle B = {110^0} \\
  \angle B + \angle C = {180^0} \\
   \Rightarrow \angle C = {70^0} \\
 \]
Hence\[{\text{ }}\angle B = {110^0}{\text{ , }}\angle C = {70^0}{\text{ , }}\angle D = {110^0}\] .

 Note: The same side interior angle theorem is only valid for parallel lines. Geometric properties of parallelogram and parallel lines should be kept in mind while solving problems like this.
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