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If the area of the $\Delta ABC$ be $\Delta$ , then the value of $b^{2}sin\ 2C+c^{2}sin\ 2B$ is
Option: 1
$3\Delta$

Option: 2
$2\Delta$

Option: 3
$4\Delta$

Option: 4
$5\Delta$

Answers (1)

best_answer

Answer: Option (c) $4\Delta$

Explanation:

Given that,

$b^{2}sin\ 2C+c^{2}sin\ 2B$

$b^{2}sin\ 2C+c^{2}sin\ 2B= b^{2}\ 2sin\ C\ cos\ C+c^{2}\ 2sin\ B\ cos\ B$

$b^{2}sin\ 2C+c^{2}sin\ 2B= 2b\ cos\ C\cdot b\ sin\ C+2c\ cos\ B\cdot c\ sin\ B$

We know that,

$\frac{a}{sin\ B}=\frac{c}{sin C}\Rightarrow b\ sin\ C=c\ sin\ B$

Thus,

$b^{2}sin\ 2C+c^{2}sin\ 2B= 2b\ cos\ C\cdot c\ sin\ B+2c\ cos\ B\cdot c\ sin\ B$

$b^{2}sin\ 2C+c^{2}sin\ 2B= 2c\ sin\ B\left ( b\ cos\ C+c\ cos\ B \right )$

We know that,

$a=b\ cos\ C+c\ cos\ B$

Therefore,

$b^{2}sin\ 2C+c^{2}sin\ 2B= 2c\ sin\ B\cdot a$ $b^{2}sin\ 2C+c^{2}sin\ 2B= 4\cdot \frac{1}{2}ac\ sin\ B$ $b^{2}sin\ 2C+c^{2}sin\ 2B= 4\Delta$

Posted by

Irshad Anwar

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