In a pentagon ABCDE, ∠ A=(2x+ 9°), ∠ B=(2x+ 1°), ∠ C=(2x−1°), ∠ D=(2x+ 5°) and ∠ E=(2x−4°). Then, value of (2 x+ 10°) is :

The correct option is 116 °.

Step 1: Understand the sum of interior angles of a polygon
The formula for the sum of the interior angles of a polygon with n sides is given by:

$\text{Sum of interior angles}=(n-2)\times 180°$

For a pentagon, the number of sides is $n=5$.

$\text{Sum of interior angles}=(5-2)\times 180°=3\times 180°=540°$

Step 2: Set up the equation for the given angles
We are given the five interior angles of pentagon ABCDE:

$\angle A=2x+9°$

$\angle B=2x+1°$

$\angle C=2x-1°$

$\angle D=2x+5°$

$\angle E=2x-4°$

Adding these five angles together, we get:

$(2x+9°)+(2x+1°)+(2x-1°)+(2x+5°)+(2x-4°)=540°$

Step 3: Solve for x
Combine the like terms:

$(2x+2x+2x+2x+2x)+(9°+1°-1°+5°-4°)=540°$

$10x+10°=540°$

$10x=540°-10°$

$10x=530°$

$x=53°$

Step 4: Calculate the required value of (2x + 10°)
Now substitute $x=53°$ into the expression $2x+10°$:

$2x+10°=2(53°)+10°$

$2x+10°=106°+10°=116°$

Thus, the value of $(2x+10°)$ is 116 °.