Question

If sin(A-B) =1/2, cos(A+B) =1/2 where 0°

Answer 1

EXPLANATION.

⇒ sin(A - B) = 1/2. - - - - - (1).

⇒ cos(A + B) = 1/2. - - - - - (2).

As we know that,

We can write equation, as

⇒ sin(A - B) = sin(30°). - - - - - (1).

⇒ cos(A + B) = cos(60°). - - - - - (2).

Now, we write equation as,

⇒ A - B = 30°. - - - - - (1).

⇒ A + B = 60°. - - - - - (2).

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

⇒ 2B = 90°.

⇒ B = 45°.

Put the value of B = 45° in equation (2), we get.

⇒ A + 45° = 60°.

⇒ A = 60° - 45°.

⇒ A = 15°.

Values of A = 15° and B = 45°.

Answer 2

STEP-BY-STEP EXPL@NATION:

.

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

$\sin(A-B) = \sin {30}^{ \circ} \: \: \: \: \: \: \: \: \: \: [ \because \: \: \: \sin {30}^{ \circ} = \frac{1}{2} ]$

$Comparing \: both \: sides,$

$A-B = {30}^{ \circ} \: \: \: ...(1)$

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

$\cos(A + B) = \cos {60}^{ \circ} \: \: \: \: \: \: \: \: \: \: [ \because \: \: \: \cos {60}^{ \circ} = \frac{1}{2} ]$

$Comparing \: both \: sides,$

$A + B = {60}^{ \circ} \: \: \: ...(2)$

$We \: have,$

• $A-B = {30}^{ \circ} \: \: \: ...(1)$
• $A + B = {60}^{ \circ} \: \: \: ...(2)$

$Add \: Eq [1] \: \And \: Eq [2],$

$\implies (A-B) + (A + B) = {30}^{ \circ} + {60}^{ \circ}$

$\implies A-B + A + B = {90}^{ \circ}$

$\implies 2 A = {90}^{ \circ}$

$\implies A = {45}^{ \circ}$

$Substitute \: this \: value \: of \: A \: in \: Eq [1],$

$\implies A-B = {30}^{ \circ} \: \: \: ...(1)$

$\implies {45}^{ \circ}-{30}^{ \circ} = B$

$\implies B = {15}^{ \circ}$

$Hence,$

• $A = {45}^{ \circ} \: \: \: \And \: \: \: B = {15}^{ \circ}$