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Hint- Use the trigonometry table to find out what sin and cos angles will give these result (only acute) ,now equate those two angles with the angles given in the question ie A+B and A-B respectively, solve these two equations to get the values of A and B.

Complete step-by-step answer:

Given that: $\sin \left( {A + B} \right) = \cos \left( {A - B} \right) = \dfrac{{\sqrt 3 }}{2}$ ----- (1)

As we know that angles A and B are both acute angles so they must lie in the first quadrant.

As we know the values of some common angles that lie in the first quadrant.

$\because \sin {60^0} = \cos {30^0} = \dfrac{{\sqrt 3 }}{2}$

Now, let us consider the parts of equation (1) separately

$

\because \sin \left( {A + B} \right) = \dfrac{{\sqrt 3 }}{2}\& \sin {60^0} = \dfrac{{\sqrt 3 }}{2} \\

\Rightarrow \sin \left( {A + B} \right) = \sin {60^0} \\

\Rightarrow \left( {A + B} \right) = {60^0}.........(2) \\

$

Similarly we solve for second part

$

\because \cos \left( {A - B} \right) = \dfrac{{\sqrt 3 }}{2}\& \cos {30^0} = \dfrac{{\sqrt 3 }}{2} \\

\Rightarrow \cos \left( {A - B} \right) = \cos {30^0} \\

\Rightarrow \left( {A - B} \right) = {30^0}.........(3) \\

$

In order to evaluate A and B let us solve equation (2) and equation (3)

First let us add equation (2) and equation (3)

\[

\Rightarrow \left( {A + B} \right) + \left( {A - B} \right) = {60^0} + {30^0} \\

\Rightarrow 2A = {90^0} \\

\Rightarrow A = \dfrac{{{{90}^0}}}{2} \\

\Rightarrow A = {45^0} \\

\]

Now let us substitute the value of in equation (2)

$

\because \left( {A + B} \right) = {60^0} \\

\Rightarrow \left( {{{45}^0} + B} \right) = {60^0} \\

\Rightarrow B = {60^0} - {45^0} \\

\Rightarrow B = {15^0} \\

$

Hence, the values of angle A and B are ${45^0}\& {15^0}$ respectively.

Note- In order to solve such problems students must remember the values of trigonometric functions for some common trigonometric angles. Also students must remember the quadrant system and where the trigonometric functions are repeating. The problem can also be done by the method of inverse. But that is a bit hard to understand for beginners.

Complete step-by-step answer:

Given that: $\sin \left( {A + B} \right) = \cos \left( {A - B} \right) = \dfrac{{\sqrt 3 }}{2}$ ----- (1)

As we know that angles A and B are both acute angles so they must lie in the first quadrant.

As we know the values of some common angles that lie in the first quadrant.

$\because \sin {60^0} = \cos {30^0} = \dfrac{{\sqrt 3 }}{2}$

Now, let us consider the parts of equation (1) separately

$

\because \sin \left( {A + B} \right) = \dfrac{{\sqrt 3 }}{2}\& \sin {60^0} = \dfrac{{\sqrt 3 }}{2} \\

\Rightarrow \sin \left( {A + B} \right) = \sin {60^0} \\

\Rightarrow \left( {A + B} \right) = {60^0}.........(2) \\

$

Similarly we solve for second part

$

\because \cos \left( {A - B} \right) = \dfrac{{\sqrt 3 }}{2}\& \cos {30^0} = \dfrac{{\sqrt 3 }}{2} \\

\Rightarrow \cos \left( {A - B} \right) = \cos {30^0} \\

\Rightarrow \left( {A - B} \right) = {30^0}.........(3) \\

$

In order to evaluate A and B let us solve equation (2) and equation (3)

First let us add equation (2) and equation (3)

\[

\Rightarrow \left( {A + B} \right) + \left( {A - B} \right) = {60^0} + {30^0} \\

\Rightarrow 2A = {90^0} \\

\Rightarrow A = \dfrac{{{{90}^0}}}{2} \\

\Rightarrow A = {45^0} \\

\]

Now let us substitute the value of in equation (2)

$

\because \left( {A + B} \right) = {60^0} \\

\Rightarrow \left( {{{45}^0} + B} \right) = {60^0} \\

\Rightarrow B = {60^0} - {45^0} \\

\Rightarrow B = {15^0} \\

$

Hence, the values of angle A and B are ${45^0}\& {15^0}$ respectively.

Note- In order to solve such problems students must remember the values of trigonometric functions for some common trigonometric angles. Also students must remember the quadrant system and where the trigonometric functions are repeating. The problem can also be done by the method of inverse. But that is a bit hard to understand for beginners.

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