Question

In a regular heptagon ABCDEFG, EH is a angle bisector of angle E . Find angle EHB.​

Answer 1

In a regular heptagon ABCDEFG, EH is a angle bisector of angle E . The angle EHB is 90.002 degree

Step-by-step explanation:

The formula for total angle of a polygon is given by $(n-2)\times180$ where n is the number of sides of the polygon.

Given that ABCDEFG is a regular heptagon and EH is a angle bisector of angle E, then the total angle of the heptagon is

$(7-2)\times180=5\times180=900~degree$

So each interior angle is $\frac{900}{7}=128.571~degree$ (given heptagon is regular).

Since EH is a angle bisector of angle E, then angle DEH and angle FEH are equal. That is $\frac{128.571}{2}=64.285~degree$.

We have to find the angle EHB. For that we consider the pentagon HBCDE.

The total angle of the pentagon is given by $(5-2)\times180=3\times180=540~degree$.

Therefore angle EHB=540-(128.517+128.571+128.571+64.285)=90.002 degree.

(the figure of the heptagon is given below)

Answer 2

Answer:

Step-by-step explanation:

The formula for total angle of a polygon is

(n-2)x 180 = total angle for  a polygon

Given that ABCDEFG is a regular heptagon and EH is a angle bisector of angle E, then the total angle of the heptagon is

(7-2)x 180 = 900°

So each interior angle is  

900 ÷ 7 =128.571

(given heptagon is regular).

Since EH is a angle bise28.571 ÷2 = 64.2855

We have to find the angle EHB. For that we consider the pentagon HBCDE.

The total angle of the pentagon is given by .

=(n-2) x 180

= 5-2 x 180ctor of angle E, then angle DEH and angle FEH are equal. That is .

1

=3 x 180

= 540°

Therefore angle EHB=The total angle of the pentagon -(each interior angle x 3 +∠ DEH and ∠FEH )

Therefore angle EHB = 540-(128.517+128.571+128.571+64.285)= 90.002 degree.

Ans= ∠EHB =90.002