Question

In the diagram
ABCD are a point in a circle
ADE and BCE are straight line
Angle DCE =56
Angle ABE =5x-16 and angle BAE=3/2x+4y
If angle DCE:angle CDE=7:8
Fine angle AEB
The value of x and y

Answer 1

Interpretation: A, B, C, D lie on a circle and E is the intersection of the lines AD and BC (so ADE and BCE are straight lines).

1) From triangle CDE:
- Given DCE : CDE = 7 : 8 and DCE = 56°, so scale factor k = 56/7 = 8.
- Thus CDE = 8k = 64°.
- Angle CED = 180° − 56° − 64° = 60°.

Since A, D, E are collinear and B, C, E are collinear, angle AEB equals angle CED (vertical/opposite angles), so
- angle AEB = 60°.

2) Use the circle: inscribed angle CDA (angle formed at D by chords DC and DA) equals angle CBA (angle at B formed by chords BC and BA) because they subtend the same arc CA. But CDA = CDE = 64° (DE is the same line as DA), so
- ABE = 64°.

Given ABE = 5x − 16, so 5x − 16 = 64 → 5x = 80 → x = 16.

3) Sum of angles in triangle ABE:
(5x − 16) + (3/2 x + 4y) + 60 = 180.
Plug x = 16: 64 + (1.5·16 + 4y) + 60 = 180 → 64 + 24 + 4y + 60 = 180
→ 148 + 4y = 180 → 4y = 32 → y = 8.

Answer:
- angle AEB = 60°
- x = 16
- y = 8.