Question

In the adjoining figure, AB = 10 cm, BC =15 cm and AD : DC = 2 : 3, then $\angle$ABC is equal to -

• A. 30º
• B. 40º
• C. 45º
• D. 110º

Answer 1

The correct option is B

40º

Clearly, $\frac{AD}{DC}$ = $\frac{2}{3}$ and $\frac{AB}{DC}$ = $\frac{10}{15}$ = $\frac{2}{3}$

So, $\frac{AD}{DC}$ = $\frac{AB}{BC}$

Thus, BD divides AC in the ratio of the other two sides.

$\therefore$ BD is the bisector of $\angle$B.

Now, $\angle$CBD = ${180}^{\circ }$ - (${130}^{\circ }$ + ${30}^{\circ }$) = ${20}^{\circ }$

$\angle$B = 2 ( $\angle$CBD)= ${40}^{\circ }$