Question

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

Answer 1

Answer:

A=30°

B=15°

.. hope this helps you ..

Answer 2

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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