Math, asked by pranjal1908, 11 months ago

if sin(A+B) = 1 and cos(A-B) = 1, 0° < A+B<_ 90°, A>_B. find A and B.​

Answers

Answered by amitkumar44481
3

AnsWer :

A = 45 and B = 45.

Solution :

We have,

 \tt\mapsto\sin ( A + B ) = 1.

 \tt\mapsto\sin ( A + B ) =  \sin90 \degree .

 \tt\mapsto ( A + B ) = 90.  \:  \:  \:  \:  \: -  (1)

\rule{90}1

 \tt\mapsto\cos ( A  -  B ) = 1.

 \tt\mapsto\cos ( A  -  B ) =  \cos 0 \degree.

 \tt\mapsto( A  -  B ) = 0. \:  \:  \:   \:  \: - (2)

Adding equation (1) and (2), We get.

\begin{tabular}{1-1}  A + B = 90. &amp; \\ A - B = 0. &amp; \\ \cline{1-1} 2A = 90. &amp; \\  \cline{1-1} \end{tabular}

 \tt\mapsto A  = 45 \degree.

Now, Putting the value of A in equation (1), We get.

 \tt\mapsto ( A + B ) = 90.

 \tt\mapsto45+ B=90.

 \tt\mapsto B = 45.

Therefore, the value of A be 45 and Value of B be 45.

\rule{200}3

Let's Verify :

We have,

 \tt\mapsto ( A + B ) = 90.

 \tt\mapsto 45 + 45  = 90.

 \tt\mapsto90 = 90.

\rule{50}1

Other equation,

 \tt\mapsto A  -  B  = 0.

 \tt\mapsto45 - 45 = 0.

 \tt\mapsto0 = 0.

Verify.

Answered by swarnpratap05
0

I hope this will help you

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