In the adjoining figure, $A B = 10 c m, B C = 15 c m$ and $A D : D C = 2 : 3,$ then $∠ A B C$ is equal to

30º

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40º

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45º

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110º

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Solution

The correct option is B
40º

Clearly, $\frac{A D}{D C} = \frac{2}{3}$ and $\frac{A B}{B C} = \frac{10}{15} = \frac{2}{3}$

So, $\frac{A D}{D C} = \frac{A B}{B C}$

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

$∴$ BD is the bisector of $∠ B .$

Now, $∠ C B D = 180^{\circ} - (130^{\circ} + 30^{\circ}) = 20^{\circ}$

$∠ B = 2 (∠ C B D) = 40^{\circ}$

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