Math, asked by sanasalmani08082004, 4 months ago

If Sin (A+2B) = 1/2 and Cos ( A +4B ) = 1/√2,A>B then find A and B​

Answers

Answered by gopalagarwal9770
55

Answer:

A=30°

B=15°

.. hope this helps you ..

Attachments:
Answered by hukam0685
0

Values are A= 15° and B=7.5°

Given:

  • sin \: (A+2B) =  \frac{1}{2}  \\ and
  • cos(A+4B) =  \frac{1}{ \sqrt{2} }  \\

To find:

  • Find the value of A and B, where A>B.

Solution:

Concept to be used:

sin \:  {30}^{ \circ}  =  \frac{1}{2}  \\

cos\:  {45}^{ \circ}  =  \frac{1}{ \sqrt{2} }  \\

Step 1:

Remove trigonometric ratios from both equations.

sin(A+2B) =  \frac{1}{2}  \\

sin(A+2B) = sin( {30}^{ \circ} ) \\

Cancel sine from both sides.

\bf A+2B = 30 ^{ \circ} ...eq1 \\

Similarly

cos(A+4B) =  {cos}45^{ { \circ} }  \\

\bf A+4B=  {45}^{ \circ} ...eq2  \\

Step 2:

Solve eq1 and eq2 for A and B.

Subtract eq1 and eq2.

A+2B = 30 ^{ \circ} \\ A+4B = 45 ^{ \circ} \\( - ) \:  \: ( - ) \:  \: ( - ) \\  -  -  -  -  -  -  -  \\  - 2B =  - 15 ^{ \circ}

 B=  \frac{15}{2}  \\

\bf B= 7.5 ^{ \circ}  \\

put the value of B in eq1

A + 2 \times  \frac{15}{2}  = 30 \\

A = 30 ^{ \circ}- 15^{ \circ} \\

\bf A = 15^{ \circ}  \\

Thus,

Values are A= 15° and B=7.5°.

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