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Question: If \(\sin \left( A-B \right)=\dfrac{1}{2},\cos \left( A+B \right)=\dfrac{1}{2}\), \({{0}^{0}}\) \(<\...

If sin⁡(A−B)=12,cos⁡(A+B)=12\sin \left( A-B \right)=\dfrac{1}{2},\cos \left( A+B \right)=\dfrac{1}{2}, 00{{0}^{0}} <<A+B≤\le 900{{90}^{0}} and A>>B , find A and B.

Explanation

Solution

Hint:In the given question, the value of sin⁡(A−B)\sin \left( A-B \right) is equal to 12\dfrac{1}{2}and we know that sin⁡30∘=12\sin {{30}^{\circ }}=\dfrac{1}{2} . Now, we get sin⁡(A−B)=sin⁡30∘\sin \left( A-B \right)=\sin {{30}^{\circ }} so A−B=300A-B={{30}^{0}} . Similarly, we can write cos⁡(A+B)=cos⁡60∘\cos \left( A+B \right)=\cos {{60}^{\circ }} because cos⁡60∘=12\cos {{60}^{\circ }}=\dfrac{1}{2} so A+B=600A+B={{60}^{0}}. Now solve A−B=300A-B={{30}^{0}} and A+B=600A+B={{60}^{0}} and find the values of A and B.

Complete step-by-step answer:
It is given in the question that,
sin⁡(A−B)=12\sin \left( A-B \right)=\dfrac{1}{2}
We know that from the trigonometric values that sin⁡30∘=12\sin {{30}^{\circ }}=\dfrac{1}{2} so in the above equation we can write:
sin⁡(A−B)=sin⁡30∘\sin \left( A-B \right)=\sin {{30}^{\circ }}
Taking sin-1 both the sides we get,
A−B=30∘A-B={{30}^{\circ }}
It is also given in the question that,
cos⁡(A+B)=12\cos \left( A+B \right)=\dfrac{1}{2}
We know that from the trigonometric values that cos⁡60∘=12\cos {{60}^{\circ }}=\dfrac{1}{2} so in the above equation we can write:
cos⁡(A+B)=cos⁡60∘\cos \left( A+B \right)=\cos {{60}^{\circ }}
Taking cos⁡−1{{\cos }^{-1}} on both the sides we get,
A+B=600A+B={{60}^{0}}
Now, we have two equations in A and B.
A−B=300……….Eq.(1)A-B={{30}^{0}} ………. Eq. (1)
A+B=600……….Eq.(2)A+B={{60}^{0}} ………. Eq. (2)
Solving above equations by elimination method we get,
Adding eq. (1) and eq. (2) will give:
2A=900 ⇒A=450 \begin{aligned} & 2A={{90}^{0}} \\\ & \Rightarrow A={{45}^{0}} \\\ \end{aligned}
Substituting this value of A in eq. (2) we get,
450+B=600 ⇒B=150 \begin{aligned} & {{45}^{0}}+B={{60}^{0}} \\\ & \Rightarrow B={{15}^{0}} \\\ \end{aligned}
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:
00<A+B≤900{{0}^{0}}<{A+B}\le {{90}^{0}} and A>BA>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.