Question

If $\sin \left( {A + B} \right) = 1$ and $\cos \left( {A - B} \right) = \dfrac{{\sqrt 3 }}{2}$, $0^\circ < A + B \leqslant 90^\circ $, $A > B$, then find the value of $A$ and $B$.

Answer 1

Hint: In this question, we are given two equations in trigonometric terms and we have been asked to find the value of $A$ and $B$. Observe the trigonometric table of values upto $90^\circ $. Compare the given equations with that table. You will further get two equations in terms of $A$ and $B$. You have two equations in two variables. Use both the equations and use the method of substitution or reduction to find their values.

Complete step-by-step solution:
We have two trigonometric equations $\sin \left( {A + B} \right) = 1$ and $\cos \left( {A - B} \right) = \dfrac{{\sqrt 3 }}{2}$. To answer this question, you must be aware about the trigonometric table.
Let us have a look at that table once.

Ratios/Angle	$0^\circ $	$30^\circ $	$45^\circ $	$60^\circ $	$90^\circ $
Sin	$0$	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$
Cos	$1$	$\dfrac{{\sqrt 3 }}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{1}{2}$	$0$

Now, we have the values of all the required trigonometric ratios. We will use them to do the question.
We have $\sin \left( {A + B} \right) = 1$. If we look at the table, we can see that $\sin 90^\circ = 1$. Therefore, we can write –
$ \Rightarrow \sin \left( {A + B} \right) = \sin 90^\circ $
On comparing we can tell, $A + B = 90$ …. (1)
We also have $\cos \left( {A - B} \right) = \dfrac{{\sqrt 3 }}{2}$. Looking at the table, we can observe that $\cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$. We can write –
$ \Rightarrow \cos \left( {A - B} \right) = \cos 30^\circ $
On comparing we can tell, $A - B = 30$ …. (2)
Now, we will use equation (1) and (2) to find the values.
Adding both the equations, we have –
$
  {\text{ }}A + B = 90 \\
  \underline {{\text{ + }}A - B = 30} \\
  {\text{ 2}}A + 0 = 120
 $
$ \Rightarrow 2A = 120$
Hence,
$ \Rightarrow A = \dfrac{{120}}{2} = 60^\circ $
Putting this in equation (1),
$ \Rightarrow 60 + B = 90$
$ \Rightarrow B = 90 - 60 = 30^\circ $
Hence, the required value of A and B is $60^\circ $ and $30^\circ $ respectively.

Note: We have to mind that, the value of sin and cos is 1 and $\dfrac{{\sqrt 3 }}{2}$ respectively at other angles also but here, we will consider only the first quadrant because it is clearly given that $0^\circ < A + B \leqslant 90^\circ $. This is where we use the given conditions.