Question

Two interior angles $A$ and $B$ of pentagon $ABCDE$ are $60^{\circ}$ and $85^{\circ}$. Two of the remaining angles, $C$ and $D$, are equal and the fifth angle $E$ is $15^{\circ}$ more than twice $C$. Find the measure of the largest angle.

Answer 1

The largest angle is E = 205°.

What is a pentagon?

• A pentagon is any five-sided polygon or 5-gon in geometry.
• A basic pentagon's internal angles add up to 540°.
• A pentagon might be straightforward or self-intersecting.
• A pentagram is a self-intersecting regular pentagon (or a star pentagon).

To find the measure of the largest angle:

The sum of interior angles of a polygon is:

• Sum = 180° × (n - 2)

Where n is the number of sides.

Here n = 5, as pentagon has 5 sides.

• Sum = 180° × 3 = 540°

Let, C = D = x

Then, E = 2x + 15 (15 more than twice C)

We can express the sum of the 5 angles as:

• 60 + 85 + x + x + 2x + 15 = 540

Simplify the left side by collecting like terms:

• 4x + 160 = 540

Subtract 160 from both sides:

• 4x + 160 - 160 = 540 - 160
• 4x = 380

Divide both sides by 4:

• 4x/4 = 380/4 = x = 95
• C = D = x = 95°
• E (2 × 95) + 15 = 205°

The 5 interior angles of the pentagon are:

• 60° + 85° + 95° + 95° + 205° = 540°

The largest angle is E = 205°.

Therefore, the largest angle is E = 205°.

Know more about a pentagon here:

brainly.com/question/25917702

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