In the figure, not drawn to scale, ZBCD is a parallelogram and ZAE is an isosceles triangle. DEBA is a straight line. ∠BCD = 100°, ∠ZBE = 57° and ∠BZA = 13°. Find ∠DZE.
3 m
In the figure, not drawn to scale, ZBCD is a parallelogram and ZAE is an isosceles triangle. DEBA is a straight line. ∠BCD = 100°, ∠ZBE = 57° and ∠BZA = 13°. Find ∠DZE.
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Ischool Pte Ltd
Date:17 May 2024
Topic:
ANGLES
Question Code:
P6.AN.NK.17051727
Student:
Class:
Level:
School:
In the figure, not drawn to scale, ZBCD is a parallelogram and ZAE is an isosceles triangle. DEBA is a straight line. ∠BCD = 100°, ∠ZBE = 57° and ∠BZA = 13°. Find ∠DZE.
(3 m)
Ischool Pte Ltd
Date:
17 May 2024
Topic:
ANGLES
Question Code:
P6.AN.NK.17051727
Student:
Class:
Level:
School:
In the figure, not drawn to scale, ZBCD is a parallelogram and ZAE is an isosceles triangle. DEBA is a straight line. ∠BCD = 100°, ∠ZBE = 57° and ∠BZA = 13°. Find ∠DZE.
In the figure, not drawn to scale, ZBCD is a parallelogram and ZAE is an isosceles triangle. DEBA is a straight line. ∠BCD = 100°, ∠ZBE = 57° and ∠BZA = 13°. Find ∠DZE.
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ISchool Pte Ltd
Date:17 May 2024
Topic:ANGLES
Question Number:P6.AN.NK.17051727-A1
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