E–ABCD has rhombus base AB=2√2 and ∠DAB=π/3. EB=EC, ∠BEC=π/2, DE=√5. F is the midpoint of BC; G is the perpendicular foot from E to DF. Show cos∠DFE=√3/4.

01 / Standard#Your turn

Official paper · jm02-2023 · 1(a) · PDF 3

Skills and prerequisite lessons

Working and explanation

BUILD THE REASONING

Hint 1

Use the median to the hypotenuse of isosceles right triangle BEC.

Hint 2

Triangles DAB and DBC are equilateral.

Worked solution

The midpoint F gives EF=BF=√2.
$EF=BC/2=\sqrt2$
In equilateral DBC, DF is an altitude.
$DF^2=DB^2-BF^2=8-2=6$
Apply the cosine rule at F.
$\cos\angle DFE=\frac{DF^2+EF^2-DE^2}{2DF\,EF}=\frac{6+2-5}{2\sqrt6\sqrt2}=\frac{\sqrt3}4$

cos∠DFE=√3/4.

Checks and common pitfalls: The cosine formula must use the angle at F, opposite DE.

Reasoning checklist · self / teacher assessment

Teaching assessment checklist, independently authored. Use the original paper for official marks.
The midpoint F gives EF=BF=√2.
$EF=BC/2=\sqrt2$
In equilateral DBC, DF is an altitude.
$DF^2=DB^2-BF^2=8-2=6$
Apply the cosine rule at F.
$\cos\angle DFE=\frac{DF^2+EF^2-DE^2}{2DF\,EF}=\frac{6+2-5}{2\sqrt6\sqrt2}=\frac{\sqrt3}4$

Think first. Reveal a hint when the class is ready.

2023 JM02