this is ques 15
and q 16.....
Mark me as brainliest
Step-by-step explanation:
ans:15
/_DBA = 54 , AS 180 - 108-18= 54 , SUM OF ANGLES OF TRIANGLE
AS ADC Is straight line so ,
/_ADB + /_ CDB = 180
/_CDB = 180 - 108= 72
therefore ,
/_ DBC = 180 - 72 - 48= 60
ans :16
/_ACB =180 - 105 - 30= 45
so , /_ ACD = 180 - 45 = 135
then /_ DEC = 180 - 135 - 30 = 15,
therefore , CA being straight line
/_AED = 180 - 15 = 165
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Answers & Comments
this is ques 15
and q 16.....
Mark me as brainliest
Step-by-step explanation:
ans:15
/_DBA = 54 , AS 180 - 108-18= 54 , SUM OF ANGLES OF TRIANGLE
AS ADC Is straight line so ,
/_ADB + /_ CDB = 180
/_CDB = 180 - 108= 72
therefore ,
/_ DBC = 180 - 72 - 48= 60
ans :16
/_ACB =180 - 105 - 30= 45
so , /_ ACD = 180 - 45 = 135
then /_ DEC = 180 - 135 - 30 = 15,
therefore , CA being straight line
/_AED = 180 - 15 = 165