In the figure, not drawn to scale, ACDE is a parallelogram and ABF is an isosceles triangle. EFCB is a straight line. ∠CDE = 101°, ∠ACF = 57° and ∠CAB = 12°. Find ∠EAF.
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In the figure, not drawn to scale, ACDE is a parallelogram and ABF is an isosceles triangle. EFCB is a straight line. ∠CDE = 101°, ∠ACF = 57° and ∠CAB = 12°. Find ∠EAF.
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Ischool Pte Ltd
Date:05 May 2024
Topic:
ANGLES
Question Code:
P6.AN.NK.17051727
Student:
Class:
Level:
School:
In the figure, not drawn to scale, ACDE is a parallelogram and ABF is an isosceles triangle. EFCB is a straight line. ∠CDE = 101°, ∠ACF = 57° and ∠CAB = 12°. Find ∠EAF.
(3 m)
Ischool Pte Ltd
Date:
05 May 2024
Topic:
ANGLES
Question Code:
P6.AN.NK.17051727
Student:
Class:
Level:
School:
In the figure, not drawn to scale, ACDE is a parallelogram and ABF is an isosceles triangle. EFCB is a straight line. ∠CDE = 101°, ∠ACF = 57° and ∠CAB = 12°. Find ∠EAF.
In the figure, not drawn to scale, ACDE is a parallelogram and ABF is an isosceles triangle. EFCB is a straight line. ∠CDE = 101°, ∠ACF = 57° and ∠CAB = 12°. Find ∠EAF.
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ISchool Pte Ltd
Date:05 May 2024
Topic:ANGLES
Question Number:P6.AN.NK.17051727-A1
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