
If \[\sin A = \dfrac{1}{{\sqrt {10} }}\]and $\sin B = \dfrac{1}{{\sqrt 5 }}$, where A and B are positive acute angles, then A + B is equal to
A.$\pi $
B.$\dfrac{\pi }{2}$
C.$\dfrac{\pi }{3}$
D.$\dfrac{\pi }{4}$
Answer
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Hint: We are given the values of sin A and sin B. So, we will find the values of cos A and cos B using the appropriate trigonometric formula and the given values of sine. Then, we will put these values in a trigonometric formula which is cos(A+B) = cos A cos B – sin A sin B to find the value of A + B.
Complete step-by-step answer:
Given: $\sin A = \dfrac{1}{{\sqrt {10} }}$
$\sin B = \dfrac{1}{{\sqrt 5 }}$
A and B are acute angles.
Now we will find the value of cos A using the formula ${\sin ^2}A + {\cos ^2}A = 1$.
Putting the value of sin A in the above formula.
${\left( {\dfrac{1}{{\sqrt {10} }}} \right)^2} + {\cos ^2}A = 1$
${\cos ^2}A = 1 - \dfrac{1}{{10}}$
${\cos ^2}A = \dfrac{{10 - 1}}{{10}}$
${\cos ^2}A = \dfrac{9}{{10}}$
$\cos A = \dfrac{3}{{\sqrt {10} }}$.
Using the same formula we will find the value of cos B.
${\sin ^2}B + {\cos ^2}B = 1$
Putting the value of sin B in the above equation.
${\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2} + {\cos ^2}B = 1$
${\cos ^2}B = 1 - \dfrac{1}{5}$
${\cos ^2}B = \dfrac{{5 - 1}}{5}$
${\cos ^2}B = \dfrac{4}{5}$
$\cos B = \dfrac{2}{{\sqrt 5 }}$
Now, we have to find values of A + B. So, we will let it equal to $\theta $.
$A + B = \theta $
Taking cos on both sides.
$\cos \left( {A + B} \right) = \cos \theta $
$\cos A\cos B - \sin A\sin B = \theta $
Putting the values in the above equation.
$\dfrac{3}{{\sqrt {10} }} \times \dfrac{2}{{\sqrt 5 }} - \dfrac{1}{{\sqrt {10} }} \times \dfrac{1}{{\sqrt 5 }} = \cos \theta $
$\dfrac{{6 - 1}}{{\sqrt {10} .\sqrt 5 }} = \cos \theta $
$\cos \theta = \dfrac{5}{{\sqrt {50} }}$
$\cos \theta = \dfrac{5}{{5\sqrt 2 }}$
$\cos \theta = \dfrac{1}{{\sqrt 2 }}$
Value of $\cos \theta $ is $\dfrac{1}{{\sqrt 2 }}$at $\dfrac{\pi }{4}$. So,
$\cos \theta = \cos \dfrac{\pi }{4}$
$\theta = \dfrac{\pi }{4}$
\[\theta \] is equal to A + B. So,
$A + B = \dfrac{\pi }{4}$.
So, option (4) is the correct answer.
So, the correct answer is “Option 4”.
Note: While finding the square root of the squared terms we have to consider both the positive and negative values. But in this case we will only consider the positive value of cos A and cos B because A and B are acute angles. So, they will lie in the first quadrant and the value of cos in the first quadrant is always positive.
Complete step-by-step answer:
Given: $\sin A = \dfrac{1}{{\sqrt {10} }}$
$\sin B = \dfrac{1}{{\sqrt 5 }}$
A and B are acute angles.
Now we will find the value of cos A using the formula ${\sin ^2}A + {\cos ^2}A = 1$.
Putting the value of sin A in the above formula.
${\left( {\dfrac{1}{{\sqrt {10} }}} \right)^2} + {\cos ^2}A = 1$
${\cos ^2}A = 1 - \dfrac{1}{{10}}$
${\cos ^2}A = \dfrac{{10 - 1}}{{10}}$
${\cos ^2}A = \dfrac{9}{{10}}$
$\cos A = \dfrac{3}{{\sqrt {10} }}$.
Using the same formula we will find the value of cos B.
${\sin ^2}B + {\cos ^2}B = 1$
Putting the value of sin B in the above equation.
${\left( {\dfrac{1}{{\sqrt 5 }}} \right)^2} + {\cos ^2}B = 1$
${\cos ^2}B = 1 - \dfrac{1}{5}$
${\cos ^2}B = \dfrac{{5 - 1}}{5}$
${\cos ^2}B = \dfrac{4}{5}$
$\cos B = \dfrac{2}{{\sqrt 5 }}$
Now, we have to find values of A + B. So, we will let it equal to $\theta $.
$A + B = \theta $
Taking cos on both sides.
$\cos \left( {A + B} \right) = \cos \theta $
$\cos A\cos B - \sin A\sin B = \theta $
Putting the values in the above equation.
$\dfrac{3}{{\sqrt {10} }} \times \dfrac{2}{{\sqrt 5 }} - \dfrac{1}{{\sqrt {10} }} \times \dfrac{1}{{\sqrt 5 }} = \cos \theta $
$\dfrac{{6 - 1}}{{\sqrt {10} .\sqrt 5 }} = \cos \theta $
$\cos \theta = \dfrac{5}{{\sqrt {50} }}$
$\cos \theta = \dfrac{5}{{5\sqrt 2 }}$
$\cos \theta = \dfrac{1}{{\sqrt 2 }}$
Value of $\cos \theta $ is $\dfrac{1}{{\sqrt 2 }}$at $\dfrac{\pi }{4}$. So,
$\cos \theta = \cos \dfrac{\pi }{4}$
$\theta = \dfrac{\pi }{4}$
\[\theta \] is equal to A + B. So,
$A + B = \dfrac{\pi }{4}$.
So, option (4) is the correct answer.
So, the correct answer is “Option 4”.
Note: While finding the square root of the squared terms we have to consider both the positive and negative values. But in this case we will only consider the positive value of cos A and cos B because A and B are acute angles. So, they will lie in the first quadrant and the value of cos in the first quadrant is always positive.
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