Question

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

• A. ${32}^0$
• B. ${37}^0$
• C. ${43}^0$
• D. ${53}^0$

Answer 1

Answer: (B)

${37}^0$

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

$(\because Sumoftwooppositinterioranglesofatriangle=exteriorangle)$

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

$⟹\angle BDC={37}^{\circ }$

$\angle BAD+\angle ADC={180}^{\circ }(sumofco-interiorangle={180}^{\circ })$

$⟹\angle ADC={90}^{\circ }$

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

$In△ABD:$

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

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

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