Mathematics

In the given figure, AB || DC, $$\angle BAD = 90^0$$, $$\angle CBD= 28^0$$ and $$\angle BCE=65^0$$. Then $$\angle ABD=?$$


ANSWER

$$37^0$$


SOLUTION
$$\angle BDC+\angle DBC={ 65 }^{ \circ  }$$

$$ (\because Sum\quad of\quad two\quad opposit\quad interior\quad angles\quad of\quad a\quad triangle=exterior\quad angle)$$

$$\implies\quad \angle BDC+{ 28 }^{ \circ  }={ 65 }^{ \circ  }$$

$$\implies\quad \angle BDC={ 37 }^{ \circ  }$$

$$ \angle BAD+\angle ADC={ 180 }^{ \circ  }\quad (sum\quad of\quad co-interior\quad angle={ 180 }^{ \circ  })$$

$$\implies\quad \angle ADC={ 90 }^{ \circ  }$$

$$ \therefore \angle ADB=\angle ADC-\angle BDC={ 90 }^{ \circ  }-{ 37 }^{ \circ  }={ 53 }^{ \circ  }$$

$$ In\triangle ABD:$$

$$ \angle ADB+\angle ABD+\angle DAB={ 180 }^{ \circ  }$$

$$\implies\quad { 53 }^{ \circ  }+\angle ABD+{ 90 }^{ \circ  }={ 180 }^{ \circ  }$$

$$\implies\quad \angle ABD={ 180 }^{ \circ  }-{ 143 }^{ \circ  }={ 37 }^{ \circ  }$$
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