Question

If sin (A − B) = 1/2, cos (A + B) = 1/2, 0° < A + B < 90°, A > B, find A and B.

Answer 1

Answer:

A = 45°

B= 15°

Step-by-step explanation:

To find A and B

If sin (A − B) = 1/2, cos (A + B) = 1/2, 0° < A + B < 90°, A > B, find A and B.

As we know that

$sin \: 30 ^{ \circ} = \frac{1}{2} \\ \\ cos\: 60 ^{ \circ} = \frac{1}{2}$

put the value of 1/2 from above in RHS of the given expressions

$sin(A-B) = sin \: 30 ^{ \circ} \\ \\ \implies \: (A-B) = 30^{ \circ} ..eq1 \\ \\ cos(A+B) = cos \: 60 ^{ \circ} \\ \\ \implies \: (A+B) = 60^{ \circ}...eq2$

now solve these equations to find the value is A and B

on adding them

$2A = 90° \\ \\ A = 45° \\ \\ B = 15°$

Hope it helps you.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$1. \: \: \displaystyle \sf{ \sin ( A - B) = \frac{1}{2} \: \: \: and \: \: \cos ( A + B) = \frac{1}{2} \: }$

$2. \: \: \sf{ {0}^{ \circ} < A + B < {90}^{ \circ}}$

TO DETERMINE

The value of A & B

CALCULATION

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

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

Again

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

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

Adding Equation (1) & Equation (2) we get

$\displaystyle \sf{ 2A = {90}^{ \circ} \: } \: \: \:$

$\implies \: \displaystyle \sf{ A = {45}^{ \circ} \: } \: \: \:$

From Equation (1)

$\displaystyle \sf{ B = {45}^{ \circ} - {30}^{ \circ} \: }$

$\implies \: \displaystyle \sf{ B = {15}^{ \circ} \: }$

RESULT

$\boxed{ \: \: \sf{ A = {45}^{ \circ} \: \: and \: \: \: \: B = {15}^{ \circ} \: \: \: }}$

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