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Mathematics
Mathematics
Problem
In the isosceles
△
A
B
C
\triangle ABC
△
A
B
C
,
∣
A
B
∣
=
∣
B
C
∣
=
8
|\mathbf{AB}| = |\mathbf{BC}| = 8
∣
AB
∣
=
∣
BC
∣
=
8
, a point
E
E
E
divides
A
B
AB
A
B
internally in the ratio
1
:
3
1:3
1
:
3
, then the cosine of the angle between
C
E
\mathbf{CE}
CE
and
C
A
\mathbf{CA}
CA
is (where,
∣
C
A
∣
=
12
|\mathbf{CA}| = 12
∣
CA
∣
=
12
) a.
−
3
8
8
-\frac{3\sqrt{8}}{8}
−
8
3
8
b.
3
8
17
\frac{3\sqrt{8}}{17}
17
3
8
c.
3
7
8
\frac{3\sqrt{7}}{8}
8
3
7
d.
−
3
8
17
-\frac{3\sqrt{8}}{17}
−
17
3
8
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