Question

If $\sin \left( A-B \right)=\dfrac{1}{2},\cos \left( A+B \right)=\dfrac{1}{2}$, ${{0}^{0}}$$<$A+B$\le$${{90}^{0}}$ and A$>$B , find A and B.

Answer 1

Hint:In the given question, the value of $\sin \left( A-B \right)$ is equal to $\dfrac{1}{2}$and we know that $\sin {{30}^{\circ }}=\dfrac{1}{2}$ . Now, we get $\sin \left( A-B \right)=\sin {{30}^{\circ }}$ so $A-B={{30}^{0}}$ . Similarly, we can write $\cos \left( A+B \right)=\cos {{60}^{\circ }}$ because $\cos {{60}^{\circ }}=\dfrac{1}{2}$ so $A+B={{60}^{0}}$. Now solve $A-B={{30}^{0}}$ and $A+B={{60}^{0}}$ and find the values of A and B.

Complete step-by-step answer:
It is given in the question that,
$\sin \left( A-B \right)=\dfrac{1}{2}$
We know that from the trigonometric values that $\sin {{30}^{\circ }}=\dfrac{1}{2}$ so in the above equation we can write:
$\sin \left( A-B \right)=\sin {{30}^{\circ }}$
Taking sin-1 both the sides we get,
$A-B={{30}^{\circ }}$
It is also given in the question that,
$\cos \left( A+B \right)=\dfrac{1}{2}$
We know that from the trigonometric values that $\cos {{60}^{\circ }}=\dfrac{1}{2}$ so in the above equation we can write:
$\cos \left( A+B \right)=\cos {{60}^{\circ }}$
Taking ${{\cos }^{-1}}$ on both the sides we get,
$A+B={{60}^{0}}$
Now, we have two equations in A and B.
$A-B={{30}^{0}} ………. Eq. (1) $
$A+B={{60}^{0}} ………. Eq. (2)$
Solving above equations by elimination method we get,
Adding eq. (1) and eq. (2) will give:
$\begin{align}
  & 2A={{90}^{0}} \\
 & \Rightarrow A={{45}^{0}} \\
\end{align}$
Substituting this value of A in eq. (2) we get,
$\begin{align}
  & {{45}^{0}}+B={{60}^{0}} \\
 & \Rightarrow B={{15}^{0}} \\
\end{align}$
Hence, the value of A = 45° and B = 15°.

Note: You must verify that the values of A and B that you are getting satisfies the condition given in the question which is:
${{0}^{0}}<{A+B}\le {{90}^{0}}$ and $A>B$
The values of A and B that we have obtained in the above solution is:
A = 45° and B = 15°
As you can see that both A and B values are lying between 0° and 90° along with that the value of A (i.e. 45°) is greater than that of B (i.e. 15°).
Hence, the values of A and B that we have obtained are satisfying the given conditions on A and B.