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IIT JEE Advanced 2010 Paper 1
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© examsnet.com
Question : 67
Total: 84
Let ABC be a triangle such that ∠ACB =
π
6
and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a =
x
2
+ x + 1, b =
x
2
– 1 and c = 2x + 1 is(are)
−
(
2
+
√
3
)
1
+
√
3
2
+
√
3
4
√
3
Validate
Solution:
Using cosine rule for ∠C, we get
√
3
2
=
(
x
2
+
x
+
x
1
)
2
+
(
x
2
−
1
)
2
−
(
2
x
+
1
)
2
2
(
x
2
+
x
+
1
)
(
x
2
−
1
)
⇒
√
3
=
2
x
2
+
2
x
−
x
2
+
x
+
1
⇒
(
√
3
−
2
)
x
2
+
(
√
3
−
2
)
x
+
(
√
+
1
)
= 0
⇒ x =
(
2
−
√
3
)
±
√
3
2
(
√
3
−
2
)
⇒ x =
−
(
2
+
√
3
)
,
1
+
√
3
⇒ x = 1 +
√
3
since (x > 0)
© examsnet.com
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