Question

In fig. ABCD is a cyclic quadrilateral. If angle BAC=50° AND ANGLE DBC=60°, THEN FIND ANGLE BCD

Answer 1

Solution :

Given That ∠BAC = 50° and ∠DBC  = 60°

In the given fig. we see that ∠BAC and ∠BDC are in the same segment

Now, In triangle BDC ,

∠BDC + ∠BCD +∠DBC = 108°(By angle sum property )

∠BAC +∠BCD + ∠DBC = 180° (  ∵∠BDC = ∠BAC = 50°) ( angle in the same segment are equal)

⇔50° +60° +∠BCD = 180°

⇔∠BCD = 180° - 110° = 70°

Therefore ∠BCD = 70°

Answer 2

$answer$

$Given \: that \: angle \: BAC=50° \:and \: DBC=60°$

$</p><p>In \: the \: given \: fig. \: we \: see \: that \: BAC and \: DBC \: are \: in \: the \: same \: segment$

$Now, \: In \: triangle \: BDC ,$

$BDC + \: BCD + \: DBC = 108°(By \: angle \: sum \: property )$

$BAC + BCD + DBC = 180° (:BDC =BAC=50°) ( \: angle \: in \: = \: the \: same \: segment \: are \: equal)$

$50° +60° +AND = 180°$

$BCD = 180° - 110° = 70°$

$Therefore \: angle \: BCD = 70°$