Question

If sin(A– B)=

2

1

and cos(A+B)=

2

1

, find A and B.​

Answer 1

C O R R E C T Q U E S T I O N :

If sin(A– B)= ¹/2 and cos(A+B)= ¹/2, then find the value of A and B.

S O L U T I O N :

Case (I),

Given,

• Sin(A - B) = ¹/2 ------(1)

We know that,

• Sin30° = ¹/2 --------(2)

From equation (1) and (2),

.°. Sin(A - B) = Sin30°

=> A - B = 30°

=> A = B + 30° ---------(3)

Case (II),

Given,

• Cos(A + B) = ¹/2 --------(4)

We know that,

• Cos60° = ¹/2 -----------(5)

From equation (4) and 5,

.°. Cos(A + B) = Cos60°

=> A + B = 60° --------------(6)

Put eqn (3) in eqn (6),

=> B + 30° + B = 60°

=> 2B = 60° - 30°

=> 2B = 30°

=> B = 15°

Now,

Put B = 15° in equation (6),

=> A + 15° = 60°

=> A = 60° - 15°

=> A = 45°

Therefore,

The value of A and B is 45° and 15° respectively.

Answer 2

Perfect Question

If sin(A−B)=1/2, cos(A+B)=1/2; ,then find A and B

Given

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

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

Find

• A =?
• B = ?

Answer

$A \: = {45}^{o} \\ and \\ B \: = {15}^{o}$

Explanation

$\sin(A-B) = \frac{1}{2} \\ \implies \sin(A-B) = {30}^{o}$

Equating the value both side

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

$\cos(A + B) = \frac{1}{2} \\ \implies \ \cos( {60}^{o} )$

Again equating the value both side

$A + B = {60}^{o} ...........(2)eq$

Add both the equation

$2A \: = {90}^{0} \\ \implies \: A = {45}^{o}$

Putting the value A in (1)

${45}^{o} + B = {60}^{o} \\ \implies \:B \: = {15}^{o}$

Hence

$A = 45^o \\ </h3><h3>B =15^o$