prove that angleA + angleB + angleC + angleD + angleE=180°
Answers
The angle A is with the arc CD of the circle. Similarly,
→ Angle B is with the arc DE of the circle.
→ Angle C is with the arc AE of the circle.
→ Angle D is with the arc AB of the circle.
→ Angle E is with the arc BC of the circle.
We know, if each arc segment of the circle is joined with the center of the circle, then the central angle made by that arc there will be the double of that angle made by that arc on the circle. Means,
→ The angle made by the arc CD at the center of the circle will be 2A.
→ The angle made by the arc DE at the center of the circle will be 2B.
→ The angle made by the arc AE at the center of the circle will be 2C.
→ The angle made by the arc AB at the center of the circle will be 2D.
→ The angle made by the arc BC at the center of the circle will be 2E.
Now if we add each of the five central angles, it'll be equal in adding each of the five arcs of the circles.
On adding the five arcs CD, DE, AE, AB and BC, we get the whole circle. So if we add the five central angles we get 360°.
Let the radius of the circle be 'r'. We have the formula "2πr × θ / 360°" to find the length of a particular arc, where θ is the central angle made by that arc in degrees. So,
CD = 2πr × 2A / 360° = 2πr × A / 180°
DE = 2πr × 2B / 360° = 2πr × B / 180°
AE = 2πr × 2C / 360° = 2πr × C / 180°
AB = 2πr × 2D / 360° = 2πr × D / 180°
BC = 2πr × 2E / 360° = 2πr × E / 180°
On adding the five arcs, we get the whole circle, i.e.,
CD + DE + AE + AB + BC = 2πr
2πr (A + B + C + D + E) / 180° = 2πr
A + B + C + D + E / 180° = 1
A + B + C + D + E = 180°
Hence Proved!