Question

Given $\sin \left( {A - B} \right) = \dfrac{1}{2}$, $\cos \left( {A + B} \right) = \dfrac{1}{2}$ and $A + B$ is acute angle. Find A and B.

Answer 1

Here we will use the basic trigonometric values of sin and cos function. First, we will convert the RHS of the given sine function into an angle of sine to find the value of $A - B$. Then we will convert the RHS of the given cosine function into an angle of cosine to find the value of $A + B$. Then we will solve these two equations to get the value of A and B.

Complete Step by step Solution:
It is given that $\sin \left( {A - B} \right) = \dfrac{1}{2}$ and $\cos \left( {A + B} \right) = \dfrac{1}{2}$.
First, we will take the equation $\sin \left( {A - B} \right) = \dfrac{1}{2}$ and we will find the value of $A - B$.
We know that the value of the sin function at $30^\circ$ is equal to $\dfrac{1}{2}$ i.e. $\sin \left( {30^\circ } \right) = \dfrac{1}{2}$. Therefore, we get
$\Rightarrow \sin \left( {A - B} \right) = \sin \left( {30^\circ } \right)$
Now the sin function gets cancelled out. Therefore, we get
$\Rightarrow A - B = 30^\circ$…………………………..$\left( 1 \right)$
Now we will take the other equation i.e. $\cos \left( {A + B} \right) = \dfrac{1}{2}$ and we will solve it to get the value of $A + B$
We know that the value of the cos function at $60^\circ$ is equal to $\dfrac{1}{2}$ i.e. $\cos \left( {60^\circ } \right) = \dfrac{1}{2}$. Therefore, we get
$\Rightarrow \cos \left( {A + B} \right) = \cos \left( {60^\circ } \right)$
Now the cos function gets cancelled out. Therefore, we get
$\Rightarrow A + B = 60^\circ$…………………………..$\left( 2 \right)$
Now we will solve equation $\left( 1 \right)$ and equation $\left( 2 \right)$ to get the value of A and B. Therefore, we will add equation $\left( 1 \right)$ and equation$\left( 2 \right)$ to get the value of A. Therefore, we get
$\begin{array}{l}A - B + \left( {A + B} \right) = 30^\circ + 60^\circ \\\ \Rightarrow 2A = 90^\circ \end{array}$
Dividing both the side by 2, we get
$\Rightarrow A = \dfrac{{90^\circ }}{2} = 45^\circ$
Now we will put the value of A in the equation $\left( 1 \right)$ to get the value of B, we get
$45^\circ - B = 30^\circ$
$\Rightarrow B = 45^\circ - 30^\circ = 15^\circ$

Hence the value of A is $45^\circ$ and the value of B is $15^\circ$.

Note:
Acute angle is the angle which is less than the 90 degree and obtuse angle is an angle whose value is greater than the 90 degree.
$\begin{array}{l}Acute\,angle < 90^\circ \\\Obtuse\,angle > 90^\circ \end{array}$
We should also know the different properties of the trigonometric function and also in which quadrant which function is positive or negative as in the first quadrant all the functions i.e. sin, cos, tan, cot, sec, cosec is positive. In the second quadrant, only the sin and cosec function are positive and all the other functions are negative. In the third quadrant, only tan and cot function is positive and in the fourth quadrant, only cos and sec function is positive.