find the values of x and y

Accepted Solution

[tex]\bold{\huge{\underline{ Solution }}}[/tex]Let consider the given triangle be ABC According to the question, AD is the median of the triangle ABC and CE also divides the triangle into two parts. At F both the lines are intersecting. Here We have ,[tex]\sf{\angle{ EAF = 25{\degree}}}[/tex][tex]\sf{\angle{ AEF = x{\degree}}}[/tex][tex]\sf{\angle{ EBD = y {\degree}}}[/tex][tex]\sf{\angle{ FDC = 90{\degree}}}[/tex][tex]\sf{\angle{ DCF = 20{\degree}}}[/tex]In triangle FDC, By using Angle sum property [tex]\sf{ {\angle}FDC + {\angle}DCF + {\angle}CFD = 180{\degree}}[/tex]Subsitute the required values, [tex]\sf{ 90{\degree} + 20{\degree} + {\angle}CFD = 180{\degree}}[/tex][tex]\sf{ 110{\degree} + {\angle}CFD = 180{\degree}}[/tex][tex]\sf{ {\angle}CFD = 180{\degree} - 110{\degree}}[/tex][tex]\sf{ {\angle}CFD = 70{\degree}}[/tex]Here. [tex]\sf{ {\angle}CFD = {\angle}EFA = 70{\degree}}[/tex]The above angles are vertically opposite angles and vertically opposite angles are equal. Now, Again by using Angle sum property but in triangle AEF [tex]\sf{ {\angle}AEF + {\angle}EFA + {\angle}FAE = 180{\degree}}[/tex]Subsitute the required values, [tex]\sf{ x{\degree} + 70{\degree} + 25{\degree} = 180{\degree}}[/tex][tex]\sf{ x{\degree} + 95{\degree}= 180{\degree}}[/tex][tex]\sf{ x{\degree} = 180{\degree} - 95{\degree}}[/tex][tex]\bold{ x = 85{\degree}}[/tex]Thus , The value of x is 85Β° Here, In the triangle ABC , CE divides the triangle into two parts So ,[tex]\sf{ {\angle}AEF + {\angle}FEB = 180{\degree}}[/tex][tex]\sf{ 85{\degree} + {\angle}FEB = 180{\degree}}[/tex][tex]\sf{ {\angle}FEB = 180{\degree} - 85{\degree}}[/tex][tex]\sf{ {\angle}FEB = 95{\degree}}[/tex]Similarly ,[tex]\sf{ {\angle}AFE + {\angle}DFE = 180{\degree}}[/tex]Straight line angles [tex]\sf{ 70{\degree} + {\angle}DFE = 180{\degree}}[/tex][tex]\sf{ {\angle}DFE = 180{\degree} - 70{\degree}}[/tex][tex]\sf{ {\angle}DFE = 110 {\degree}}[/tex]In triangle ABC, EBDF is forming quadrilateral We know that, Sum of all the angles of quadrilateral is equal to 360Β° That is ,[tex]\sf{ {\angle}FEB + {\angle}EBD + {\angle}BDF + {\angle} DFE = 360{\degree}}[/tex]Subsitute the required values, [tex]\sf{ 95{\degree} + y{\degree} + 90{\degree} + + 110{\degree} = 360{\degree}}[/tex][tex]\sf{ y{\degree} + 295{\degree}= 180{\degree}}[/tex][tex]\sf{ y {\degree} = 360{\degree} - 295{\degree}}[/tex][tex]\bold{ y = 65{\degree}}[/tex]Hence, The value of x and y is 70Β° and 65Β° .