Question

In the figure (not drawn to scale),$DAE,CBH{\text{ and }}ACG$ are straight lines. $DAE\parallel CBH\parallel FG$ and $CA = CB$ . Find the measure of angle $x{\text{ and }}y$ respectively.
I.${70^o},{35^o}$
II.${110^o},{145^o}$
III.${110^o},{35^o}$
IV.${140^o},{20^o}$

Answer 1

Hint: The concept of alternate angles, corresponding angles, supplementary angles and co-interior angles are used in the question.

Complete step-by-step answer:
The figure of the question is shown below. The value of ${x^o}{\text{ and }}{y^o}$are to be determined.

The line $FG$ is parallel to $DAE$ and it is cut by a transversal $ACG$ . Then by the definition, $\angle FGA$ and $\angle DAG$ are Co-interior angles. Their sum is equal to${180^o}$.
 $\angle FGA + \angle DAG = {180^o} \cdots \left( 1 \right)$
Substitute the value of $\angle FGA = {70^o}{\text{ and }}\angle {\text{DAG = }}{x^o}$in equation (1),
$
  {70^o} + {x^o} = {180^o} \\
  {x^o} = {180^o} - {70^o} \\
  {x^o} = {110^o} \\
 $

$\angle FGA{\text{ and }}\angle GCB$ are alternate interior angles. So, $\angle FGA{\text{ = }}\angle GCB$. Therefore, $\angle GCB = {70^o}$.
$\angle GCB{\text{ and }}\angle ACB$ are supplementary angles . Their sum is equal to.
$\angle GCB + \angle ACB = {180^o} \cdots \left( 2 \right)$
Substitute the value of $\angle GCB = {70^o}$ in equation (2)
$
  {70^o} + \angle ACB = {180^o} \\
  \angle ACB = {180^o} - {70^o} \\
  \angle ACB = {110^o} \cdots \left( 3 \right) \\
 $
In triangle $ACB$ , the sides $CA$ and $CB$ are equal. Therefore, the $\Delta ACB$ is an isosceles triangle.
In $\Delta ACB$ using the angle sum property of triangles.
$\angle ACB + \angle CAB + \angle CBA = {180^o} \cdots \left( 4 \right)$
Substitute the value of $\angle ACB = {110^o}$ from equation (3) in equation (4),
${110^o} + \angle CAB + \angle CBA = {180^o} \cdots \left( 5 \right)$

$\angle CBA = \angle CAB$ are equal . Then equation (5) can be written as,
\[
  {110^o} + \angle CBA + \angle CBA = {180^o} \\
  2\angle CBA = {180^o} - {110^o} \\
  \angle CBA = \dfrac{{{{70}^o}}}{2} \\
  \angle CBA = {35^o} \\
 \]

Now, $\angle CBA$ and $\angle ABH$are supplementary angles. Their sum is equal to ${180^o}$
$\angle CBA + \angle CBH = {180^o} \cdots \left( 6 \right)$
Substitute the value of $\angle CBA = {55^o}$ in equation (6),
$
  {35^o} + \angle CBH = {180^o} \\
  \angle CBH = {180^o} - {35^o} \\
  \angle CBH = {145^o} \\
 $
Thus, the value of ${x^o} = {110^o}$ and${y^o} = {145^o}$.
So, the correct answer is “Option B”.

Note: The important thing in this question is the concept of types of angles formed when the straight line is intersected by a transversal.
For instance in the figure, Line $L$ and $M$ are parallel to each other. They are intersected by a transversal line $N.$

${x^o}{\text{ and }}{a^o}$ are co-interior angles ( their sum is equal to ${180^o}$ ) .
The pair of ${x^o}{\text{ and }}{y^o},{z^o}{\text{ and }}{a^o}$ are alternate interior angles (they are equal to each other).
The pair of ${a^o}{\text{ and }}{y^o}$ , are a pair of supplementary angles (their sum is equal to ${180^o}$).
The pair of ${b^o}{\text{ and }}{y^o}$ are corresponding angles.