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  • Help me solve this If in a triangle ABC, ÐC =90°, then the maximum value of sinA sinB is

 If in a triangle ABC, ÐC =90°, then the maximum value of sinA sinB is 

  • Option 1)

    \frac{1}{2}

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    None of these

 

Answers (1)

best_answer

As we learnt

 

Subtraction Formulae -

 

\cos \left ( A-B \right )= \cos A\cos B+\sin A\sin B

- wherein

A and B are two angles.

 

 

 

sinA sinB =\frac{1}{2} \times 2\sin A\sin B        =\frac{1}{2}\left[ {\cos (A - B) - \cos (A + B)} \right]

                        =\frac{1}{2}\left[ {\cos (A - B) - \cos 90^\circ } \right]= \frac{1}{2}\cos (A - B) £ \frac{1}{2}

                        Þ Maximum value of sinA sinB =\frac{1}{2}.

                        Hence (A) is the  correct answer.

 


Option 1)

\frac{1}{2}

Option 2)

1

Option 3)

2

Option 4)

None of these

Posted by

Himanshu

View full answer

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