please solve guys urgent
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as u know the given figure is a rhombus...
so triangle DOA is congurent to triangle COB..
and triangle DOC is congurent to triangle AOB..
the figure is a rhombus so, its diagonals bisected perpendicularly..
so,∠DOC=∠DOA=∠AOB=∠BOC=90°
now, as you know ...
sum of angles of a triangle is 180°..
so, in triangle AOB,
∠OAB+∠ABO+∠AOB = 180°..
so, x =60°..
now as you know..
sum of angles of a quadrilateral is 360°..
∠CDA+∠DAB+∠ABC+∠BCD = 360°
so, y can be any angle..
so triangle DOA is congurent to triangle COB..
and triangle DOC is congurent to triangle AOB..
the figure is a rhombus so, its diagonals bisected perpendicularly..
so,∠DOC=∠DOA=∠AOB=∠BOC=90°
now, as you know ...
sum of angles of a triangle is 180°..
so, in triangle AOB,
∠OAB+∠ABO+∠AOB = 180°..
so, x =60°..
now as you know..
sum of angles of a quadrilateral is 360°..
∠CDA+∠DAB+∠ABC+∠BCD = 360°
so, y can be any angle..
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