Question

in the given figure angle A = angle D =90° angle C = 48° BE is the bisector of angle B. AD and BE intersect at M. Calculate angle BMD and angle AEM​

Answer 1

Step-by-step explanation:

First in ABC,

angle ABC + BCA + ACB = 180°

therefore ABC + 48° + 90° = 180°

angle ABC = 42° ... ( 1 )

as given be is bisector of angle ABC,

angle MBD + MDA = 42°

therefore both MDB & MDA = 21°

In MBD,

angle MBD + BDM + DMB = 180°

21° + 90° + DMB = 180°

angle DMB = 69° ...( 2 )

angle DMB = angle EMA ( vertically opp angles)

therefore angle EMA = 69° ( by (2) )

In CDA,

angle CDA + DCA + CAD = 180°

therefore 90° + 48° + CAD = 180°

angle CAD = 42° ...( 3 )

in EMA,

angle EMA + MAE+ AEM = 180°

therefore 69° + 42° + AEM = 180°

angle AEM = 69°