Question

One angle of a pentagon is ${140}^{\circ }$. If the remaining angles are in the ratio $4:2:3:1$; find the smallest and the largest angles.

• A. ${50}^{\circ }$and ${150}^{\circ }$
• B. ${30}^{\circ }$and ${170}^{\circ }$
• C. ${40}^{\circ }$and ${160}^{\circ }$
• D. None of these

Answer 1

The correct option is C ${40}^{\circ }$and ${160}^{\circ }$

One angle of a pentagon is ${140}^o$. If the remaining angles are in the ratio 4:2:3:1.
Let remaining angles are $=4x,2x,3x,1x$
Sum of all interior angles of pentagon = ${180}^o(n-2)={180}^o(5-2)={540}^o$
Now,
${140}^o+4x+2x+3x+1x={540}^o$
$10x={400}^o$
$x={40}^o$
All remaining angles are $=4x=4\times {40}^o={160}^o$
$=2x=2\times {40}^o={80}^o$
$=3x=3\times {40}^o={120}^o$
$=1x=1\times {40}^o={40}^o$
Smallest angle is ${40}^o$ and Largest angle is ${160}^o$.