Question

Луч СD делит развёрнутый угол AСB на два равных угла ACD и DCB. Луч CF делит пополам угол ACD, CE делит пополам угол DCB. Найдите градусную меру ∠FCE. Выполните рисунок, запишите подробное решение с пояснениями.

Answer 1

АВС- 180(потому что развернутый угол)

ДС делит  АВС его на два равных угла

180/2=90(тогда АСД=ДСВ)

Ещё луч СF  делит угол АСД пополам

90\2=45

И луч СЕ делит угол ДСВ пополам

90/2=45

Найдем угол FCE

Answer 2

Ответ:

Ответ : 72°, 108°

Пошаговое объяснение:

Пусть х градусов - градусная мера угла DBC, тогда градусная мера угла ABD - 1,5 * х градусов. Нам известно, что луч BD делит развёрнутый угол ABC на два угла ABD и DBC, а градусная мера развернутого угла равна 180 градусов. Составляем уравнение:

х + 1,5 * х = 180;

х * (1 + 1,5) = 180;

х * 2,5 = 180;

х = 180 : 2,5;

х = 1800 : 25;

х = 72 градуса - градусная мера угла DBC;

2) 1,5 * 72 = 108 градусов - градусная мера угла ABD.

Ответ: 72 градуса; 108 градусов.

Answer 3

Problem Analysis

We are given that ray CD divides the angle ∠ACB into two equal angles, ∠ACD and ∠DCB. We are also given that ray CF bisects angle ∠ACD, and ray CE bisects angle ∠DCB. We need to find the measure of angle ∠FCE.

Solution

To find the measure of angle ∠FCE, we can use the fact that the sum of the angles in a triangle is 180 degrees. Let's break down the solution step by step:

1. Draw a diagram: - Draw a line segment AB and mark point C on it. - Draw ray CD such that it divides angle ∠ACB into two equal angles, ∠ACD and ∠DCB. - Draw ray CF such that it bisects angle ∠ACD. - Draw ray CE such that it bisects angle ∠DCB. - Label the intersection point of rays CF and CE as point F.

2. Identify known angles: - ∠ACD and ∠DCB are equal because ray CD divides angle ∠ACB into two equal angles. - ∠ACD is bisected by ray CF, so ∠ACF and ∠FCD are equal. - ∠DCB is bisected by ray CE, so ∠DCB and ∠DCB are equal.

3. Use the fact that the sum of angles in a triangle is 180 degrees: - In triangle CEF, we have ∠ACF, ∠FCE, and ∠FCD. - The sum of these angles must be 180 degrees.

4. Set up the equation: - ∠ACF + ∠FCE + ∠FCD = 180 degrees.

5. Substitute the known angles: - Since ∠ACF and ∠FCD are equal to ∠ACD, we can substitute ∠ACD for both ∠ACF and ∠FCD. - ∠ACD + ∠FCE + ∠ACD = 180 degrees.

6. Simplify the equation: - 2∠ACD + ∠FCE = 180 degrees.

7. Solve for ∠FCE: - ∠FCE = 180 degrees - 2∠ACD.

8. Substitute the measure of ∠ACD: - Since ∠ACD is half of ∠ACB, we can substitute ∠ACB/2 for ∠ACD. - ∠FCE = 180 degrees - 2(∠ACB/2).

9. Simplify the equation: - ∠FCE = 180 degrees - ∠ACB.

Therefore, the measure of angle ∠FCE is 180 degrees minus the measure of angle ∠ACB.

Conclusion

In this problem, we were given that ray CD divides the angle ∠ACB into two equal angles, ∠ACD and ∠DCB. Ray CF bisects angle ∠ACD, and ray CE bisects angle ∠DCB. We found that the measure of angle ∠FCE is equal to 180 degrees minus the measure of angle ∠ACB.