In a quadrilateral ABCD, the angles A, B, C, and D are in the ratio 2:3:3:4. Find the measure of the smaller angle.

Given, Angles of quadrilateral A, B, C, and D are in the ratio 2:3:3:4.

Let, $\angle \mathrm{A}$ = 2x.

 Therefore, $\angle \mathrm{B}$ = 3x, $\angle \mathrm{C}$ = 3x, $\angle \mathrm{D}$ = 4x

 Since the sum of angles of a quadrilateral is equal to 360 degrees.

Therefore, $\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}={360}^{\mathrm{o}}$

$\Rightarrow$${\text{2x + 3x + 3x + 4x = 360}}^{\mathrm{o}}$

$\Rightarrow {\text{12x = 360}}^o$

$\Rightarrow \mathrm{x}={30}^{\mathrm{o}}$

Therefore, the smallest angle = 2x = ${60}^o$