Question

plz solve this fast......

Answer 1

hope its helpful. ....

Answer 2

HERE YOUR ANSWER:

GIVEN:- AB=AO=OB=OD=OC

CONSTRUCTION:- JOIN D TO B (shown in above picture)

TO FIND:- angleAMB

FINDING:- angleBDA= 1/2 of angleAOB (angle subtended to centre is double to angle subtended to arc)...............(1)

In triangle AOB
AB=OB=OA
so,Triangle AOB is an equilateral triangle.
Then,
all angle will be 60 degree.
angleAOB=60degree

angleBDA=1/2×angleAOB ( according to (1) )
angleBDA=1/2×60
angleBDA=30 degree

angleCBD=1/2×angleDOC (angle subtended to centre is double to angle at arc.)
angleCBD=1/2×180
angleCBD=90 degree

angleCBD+angleDBM=180 degree (Linear pair)

angleDBM=180-90
angleDBM=90 degree

IN TRIANGLE MBD

angleDBM+angleBDM+angleDMB=180 (A.S.P of triangle)

90+30+angleDMB=180
120+angleDMB=180
angleDMB=180-120
angleDMB=60 degree


So, angleDMB=60 degree

NOTE: IN THEORAM WHERE I TYPED THAT "angle subtended to centre is double to angle subtended at arc" but i write 1/2×COD and 1/2×BOA because when we write 2×BDA and 2×CBD, It will change into 1/2×COD,1/2×BOA.

BOA and COD is angle subtended at centre.



HOPE IT HELPS YOU!