Question

angle DBC = 70 degree and angle CAB =30 degree , find angle BCD.​

Answer 1

$\rm \underline{In \: quadrilateral}$

$\rm{In \: a \: ΔCDB \: and \: ΔBAC,}$

$\rm{∠ CDB = ∠ BAC = 30° ...(1)}$

$\rm { [ Angles \: in \: the \: same \: segment \: of \: a \: circle \: are \: equal ] }$

$\rm{∠ DBC = 70° ...(2)}$

$\rm{In \: Δ BCD,}$

$\rm{∠ BCD + ∠ DBC + ∠ CDB = 180°}$

$\rm{[Sum \: of \: all \: the \: angles \: of \: a \: triangle \: is \: 180°] }$

$\rm{⇒ ∠ BCD + 70° + 30° = 180° [ from (1) and (2) ]}$

$\rm{⇒ ∠ BCD +100° =180°}$

$\rm{⇒ ∠ BCD =180° - 100°}$

$\rm{⇒ ∠ BCD = 80°}$

Answer 2

Answer:

your answer is in the attachment

Step-by-step explanation:

HOPE THIS HELPS YOU

PLEASE MARK AS BRAINLIEST