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JEE Main 23 Jan 2025 Shift 1 Paper
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Section:
Mathematics
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© examsnet.com
Question : 12
Total: 75
Let the
arc
‌
A
‌
C
of a circle subtend a right angle at the centre
O
. If the point
B
on the are
A
C
, divides the arc
A
C
such that
‌
‌
length of
‌
arc
‌
A
‌
B
‌
length of
‌
arc
‌
B
‌
C
=
‌
1
5
, and
→
O
C
=
α
→
O
A
+
β
→
O
B
, then
α
+
√
2
(
√
3
−
1
)
β
is equal to
[23 Jan 2025 Shift 1]
5
√
3
2
+
√
3
2
√
3
2
−
√
3
Validate
Solution:
Let the radius be
r
B
divides the arc in
1
:
5
, so
A
B
would subtend an angle of
15
∘
at origin
O
and similarly,
B
C
would subtend an angle of
75
∘
at origin.
→
O
A
=
→
a
,
→
O
B
=
→
b
,
→
O
C
=
→
c
→
b
=
r
(
cos
‌
75
∘
)
^
i
+
r
(
s
i
n
‌
75
∘
)
^
j
→
b
=
(
‌
√
3
−
1
2
√
2
)
→
c
+
(
‌
√
3
+
1
2
√
2
)
→
a
2
√
2
→
b
=
(
√
3
−
1
)
→
c
+
(
√
3
+
1
)
→
a
‌
→
O
C
=
−
‌
(
√
3
+
1
)
√
3
−
1
∣
α
→
O
A
+
‌
2
√
2
√
3
−
1
∣
β
→
O
B
‌
α
+
√
2
(
√
3
−
1
)
β
=
2
−
√
3
© examsnet.com
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