Question

Find the value of $ x $ in the given figure given below:
 

A. $ {118^ \circ } $
B. $ {20^ \circ } $
C. $ {72^ \circ } $
D. $ {223^ \circ } $

Answer 1

Hint: In this question we have to find the value of $ x $ . To find the value of $ x $ we will use the sum of all interior angles of a triangle property and straight angle or straight-line definition. By using these properties, we can find the value of $ x $ .
The sum of all interior angles of the triangle is $ {180^ \circ } $ .
The angle of the straight line is $ {180^ \circ } $ .

Complete step-by-step answer:
Consider the given figure. Draw a straight line from a point D which bisects the line $ AB $ . Let name the point as E which bisects the line $ AB $ .
Using the property “the sum of all interior angles of a triangle is $ {180^ \circ } $ ”
Now consider the $ \vartriangle ACE $
The sum of all interior angles is,
 $ {48^ \circ } + {30^ \circ } + a = {180^ \circ } $
 $ \Rightarrow {78^ \circ } + a = {180^ \circ } $
 $ \Rightarrow a = {180^ \circ } - {78^ \circ } $
 $ \Rightarrow a = {102^ \circ } $
Therefore the $ \left| \!{\underline {\,
  E \,}} \right. $ in the $ \vartriangle ACE $ is $ {102^ \circ } $ .
      

Now consider the straight line $ AB $ where E is the midpoint of $ AB $ . We know that the angle of the straight line is $ {180^ \circ } $ . We know value of the $ \left| \!{\underline {\,
  {AEC} \,}} \right. $ is $ {102^ \circ } $ .
Therefore, we have $ \left| \!{\underline {\,
  E \,}} \right. = \left| \!{\underline {\,
  {AEC} \,}} \right. + \left| \!{\underline {\,
  {CEB} \,}} \right. $
 $ \Rightarrow \left| \!{\underline {\,
  E \,}} \right. = {102^ \circ } + y $
 $
   \Rightarrow {180^ \circ } = {102^ \circ } + y \\
   \Rightarrow y = {180^ \circ } - {102^ \circ } \;
  $
 $ \Rightarrow y = {78^ \circ } $
Now let us consider the $ \vartriangle DEB $
The sum of all interior angles is,
 $ {40^ \circ } + {78^ \circ } + z = {180^ \circ } $
 $ \Rightarrow {118^ \circ } + z = {180^ \circ } $
 $ \Rightarrow z = {180^ \circ } - {118^ \circ } $
 $ \Rightarrow z = {62^ \circ } $
Therefore $ \left| \!{\underline {\,
  {EDB} \,}} \right. $ IS $ {62^ \circ } $ .
Now consider the straight line $ EC $ where D is the midpoint of $ EC $ . We know that the angle of the straight line is $ {180^ \circ } $ . We know value of the $ \left| \!{\underline {\,
  {DEB} \,}} \right. $ is $ {62^ \circ } $ .
Therefore, we have $ \left| \!{\underline {\,
  D \,}} \right. = \left| \!{\underline {\,
  {EDB} \,}} \right. + \left| \!{\underline {\,
  {CDB} \,}} \right. $
 $ \Rightarrow \left| \!{\underline {\,
  D \,}} \right. = {62^ \circ } + b $
 $
   \Rightarrow {180^ \circ } = {62^ \circ } + b \\
   \Rightarrow b = {180^ \circ } - {62^ \circ } \;
  $
 $ \Rightarrow b = {118^ \circ } $
Therefore, the value of $ x $ is $ {118^ \circ } $ .
So, the correct answer is “Option A”.

Note: Candidates should be careful in determining the angles which are unknown. Here use these properties. The sum of all interior angles of the triangle is $ {180^ \circ } $ and the angle of the straight line is $ {180^ \circ } $ . Or we can use another property: the sum of interior angles is equal to the opposite exterior angle.