Question

In the adjoining figure $\angle ABD={132}^o$ & $\angle EAC={120}^o$. Prove that $AB>AC$

Answer 1

$\angle ABC+\angle ABC={180}^{\circ }$

$\angle ABC=180-132={48}^{\circ }$

$\angle EAC+\angle BAC={180}^{\circ }$

$\angle BAC={180}^{\circ }-{120}^{\circ }={60}^{\circ }$

$\angle ABC+\angle BCA+\angle CAB={180}^{\circ }$

$\angle BCA=180-60-48={72}^{\circ }$

$\angle BCA$ is largest angle.

$\angle ABC$ is smallest angle

$\Rightarrow$ sides opposite to largest angle are large (AB)

sides opposite to smallest angle are small (AC)

$\therefore ABC>AC$