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Manipal Entrance Test 2015 Math Paper

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Question : 23 of 70

Marks: +1, -0

In a ΔABC, if

\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix}=0

\sin^2 A + \sin^2 B + \sin^2 C

is equal to

$\frac{3\sqrt{3}}{2}$
$\frac{9}{4}$
$\frac{5}{4}$
$2$

Solution:

We have,

\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix}=0

\Rightarrow \begin{vmatrix} 1 & a & b \\ 0 & c-a & a-b \\ 0 & b-c & c-b \end{vmatrix}=0

[applying

R_2 \Rightarrow R_2 - R_1

and

R_3 \rightarrow R_3 - R_1 ]

\Rightarrow (c-a)(c-b)+(a-b)^2=0

\Rightarrow a^2+b^2+c^2

-ab-bc-ca=0

\Rightarrow 2a^2+2b^2+2c^2

-2ab-2bc-2ca=0

\Rightarrow (a-b)^2+(b-c)^2+(c-a)^2=0

\Rightarrow a=b=c

∴ ΔABC is an equilateral triangle.

\Rightarrow A=B=C=\frac{\pi}{3}

\therefore \sin^2 A+\sin^2 B+\sin^2 C

=3\sin^2 \frac{\pi}{3}=\frac{9}{4}

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