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BITSAT 2015 Paper

Section: Mathematics

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Question : 97 of 150

Marks: +1, -0

In a ΔABC, the lengths of the two larger sides are 10 and 9 units, respectively. If the angles are in AP, then the length of the third side can be

5 ± $\sqrt{6}$
$3\sqrt{3}$
5
None of these

Solution:

Let A, B and C be the three angles of ΔABC and

Let a = 10 and b = 9

It is given that the angles are in AP.

∴ 2B = A + C on adding B both the sides, we get

3B = A + B + C

⇒ 3B = 180° ⇒ B = 60°

Now, we know cos B =

\frac{a^2+c^2-b^2}{2ac}

⇒ cos 60° =

\frac{10^2+c^2-9^2}{2 \times 10 \times c}

⇒

\frac{1}{2}

=

\frac{100+c^2-81}{20c}

⇒

c^2

- 10x + 19 = 0 ⇒ c = 5 ±

\sqrt{6}

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