Question

From the given figure find $\angle DBC$

• A. ${30}^{\circ }$
• B. ${110}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is C

${60}^{\circ }$

In the given figure the polygon has 4 sides. So it is a quadrilateral.

The sum of interior angles of a quadrilateral $={360}^{\circ }$ ($\because$ Sum of interior angles $=(n-2){180}^{\circ }$)

$\Rightarrow {60}^{\circ }+{70}^{\circ }+x^{\circ }+5x^{\circ }+4x+{30}^{\circ }={360}^{\circ }$

$\Rightarrow 10x+{160}^{\circ }={360}^{\circ }$

$\Rightarrow 10x={200}^{\circ }\Rightarrow x={20}^{\circ }$

Now BDC forms a triangle. Since sum of interior angles of a triangle $={180}^{\circ }$

$\Rightarrow x+5x+\angle DBC={180}^{\circ }$

$\Rightarrow 20+100+\angle DBC={180}^{\circ }$ ($\because x={20}^{\circ }$)

$\Rightarrow \angle DBC={180}^{\circ }-{120}^{\circ }$

$={60}^{\circ }$