If $ \sin (\mathbf{A}-\mathbf{B})=\frac{1}{2} $ and $ \cos(\mathbf{A}+\mathbf{B})=\frac{1}{2}, $$ 0^{\circ}<\mathbf{A}+\mathbf{B} \leq 90^{\circ}, \mathbf{A}>\mathbf{B} $ find $ \mathbf{A} $ and $ \mathbf{B} $.

Given:

\( \sin (\mathbf{A}-\mathbf{B})=\frac{1}{2} \) and  \( \cos(\mathbf{A}+\mathbf{B})=\frac{1}{2}, \)\( 0^{\circ}<\mathbf{A}+\mathbf{B} \leq 90^{\circ}, \mathbf{A}>\mathbf{B} \)

To do:

We have to find $A$ and $B$.

Solution:  

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

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

$\Rightarrow  A-B=30^{\circ}$......(i)

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

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

$\Rightarrow A+B=60^{\circ}$

$\Rightarrow  A=60^{\circ}-B$........(ii)

Substituting (ii) in (i), we get,

$60^{\circ}-B-B=30^{\circ}$

$\Rightarrow  2B=30^{\circ}$

$\Rightarrow  B=\frac{30^{\circ}}{2}$

$\Rightarrow  B=15^{\circ}$

$\Rightarrow  A=60^{\circ}-15^{\circ}=45^{\circ}$

The values of $A$ and $B$ are $45^{\circ}$ and $15^{\circ}$ respectively.  

Tutorialspoint

Updated on: 10-Oct-2022

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