Question

in the figure angle BAD =3angleDBA, find angleCBA, angleDBC and angleABC

Answer 1

HEYA USER!

HERE IS THE ANSWER FOR YOUR QUESTION :

ANGLE BAD +ANGLE DBA + ANGLE BDA =180

3 ANGLE DBA +ANGLE DBA + 104 = 180

4 ANGLE DBA= 180-104

4 ANGLE DBA = 76

ANGLE DBA = 76/4

ANGLE DBA = 19

ANGLE ADB + BDC = 180

104 + ANGLE BDC =180

ANGLE BDC = 76

ANGLE DBC + ANGLE DCB + BDC = 180

ANGLE DBC + 65 + 76 = 180

ANGLE DBC = 180 - 141

ANGLE DBC = 39

I HOPE IT HELPS YOU!

Answer 2

HEY MATE HERE IS UR ANSWER ( I AM WRITING IN SHORT )


GIVEN THAT A = 3B

NOW IN TRIANGLE ABD

A+ B + D = 180. ( ANGLE SUM PROPERTY )

PUTTING VALUE

A= 3B , D = 104

WE HAVE
3B + B + 104 = 180

4B = 180 - 104

4B = 76

B = 19
THERE FOR

A = 57

B = 19


NOW IN TRIANGLE DBC

ANGLE ADB AND ANGLE CDB


104 + CDB = 180. ( LINEAR PAIRS )


CDB = 76

THERE FOR NOW


ANGLE CDB + DCB + DBC = 180

NOW PUTTING ALL VALUES WE HAVE

76 +65 + DBC = 180

DBC = 39

NOW ANGLE ABC = DBC + ABD

WE HAVE

39 + 19 = 58


HOPE IT IS HELPFUL IF YES PLEASE MARK AS BRAINLIEST