Question

Question 34
The angle between the two altitudes of a parallelogram through the same vertex of an obtuse angle of the parallelogram is ${30}^{\circ }$. The measure of the obtuse angle is
a) ${100}^{\circ }$
b) ${150}^{\circ }$
c) ${105}^{\circ }$
d) ${120}^{\circ }$

Answer 1

Let EC and FC be altitudes and $\angle ECF={30}^{\circ }$.

Let $\angle EDC=x=\angle FBC$
So, $\angle ECD=90-x^{\circ }and\angle BCF={90}^{\circ }-x$
So, by property of the parallelogram,
$\angle ADC+\angle DCB={180}^{\circ }\angle ADC+(\angle ECD+\angle ECF+\angle BCF)={180}^{\circ }\Rightarrow x+{90}^{\circ }-x+{30}^{\circ }+{90}^{\circ }-x={180}^{\circ }\Rightarrow -x={180}^{\circ }-{210}^{\circ }=-{30}^{\circ }\Rightarrow x={30}^{\circ }$
Hence, $\angle DCB={30}^{\circ }+{60}^{\circ }+{60}^{\circ }={150}^{\circ }$.