
If $\sin {\text{3}}A$ =$\cos \left( {A - {{26}^ \circ }} \right)$ where $3A$ is an acute angle, find the value of $A$.
Answer
601.8k+ views
Hint: In order to solve this question we have to convert $\sin $ in terms of $\cos {\text{ }}\left[ {{\text{As }}\sin \theta = \left( {\cos {{90}^ \circ } - \theta } \right)} \right]$. By doing so we will get both sides in terms of cosine.
Complete step-by-step answer:
Thus, it will be easier to solve this question.
Complementary Angles- Two angles are said to be complementary if their sum is equal to the right angle.
In this question we have given,
$\sin {\text{3}}A$ =$\cos \left( {A - {{26}^ \circ }} \right)$
In this question, we have to find out the value of $A$ .
As we know that
$\sin x = \cos ({90^ \circ } - x)$
So, by using the identity we can say ,
$\sin 3A = \cos \left( {{{90}^ \circ } - 3A} \right) = \cos \left( {A - {{26}^ \circ }} \right)$
Since, $3A$ is an acute angle we can say that ,
$\cos \left( {{{90}^ \circ } - 3A} \right) = \cos \left( {A - {{26}^ \circ }} \right)$
As we have $\cos $ on both sides , by eliminating cos we will get,
$
{90^ \circ } - 3A = A - {26^ \circ } \\
\Rightarrow 4A = {116^ \circ } \\
\Rightarrow {\text{ }}A = \dfrac{{{{116}^ \circ }}}{4} \\
\Rightarrow {\text{ }}A = {29^ \circ } \\
$
Note: Whenever we face such types of questions, the key concept is that we must covert sin in terms of cos or vice versa. It is clearly visible that here $3A$ represents an acute angle. First we will use the identity $\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$ then eliminate cos (or sin) then by simplifying the equations we will get our required answer.
Complete step-by-step answer:
Thus, it will be easier to solve this question.
Complementary Angles- Two angles are said to be complementary if their sum is equal to the right angle.
In this question we have given,
$\sin {\text{3}}A$ =$\cos \left( {A - {{26}^ \circ }} \right)$
In this question, we have to find out the value of $A$ .
As we know that
$\sin x = \cos ({90^ \circ } - x)$
So, by using the identity we can say ,
$\sin 3A = \cos \left( {{{90}^ \circ } - 3A} \right) = \cos \left( {A - {{26}^ \circ }} \right)$
Since, $3A$ is an acute angle we can say that ,
$\cos \left( {{{90}^ \circ } - 3A} \right) = \cos \left( {A - {{26}^ \circ }} \right)$
As we have $\cos $ on both sides , by eliminating cos we will get,
$
{90^ \circ } - 3A = A - {26^ \circ } \\
\Rightarrow 4A = {116^ \circ } \\
\Rightarrow {\text{ }}A = \dfrac{{{{116}^ \circ }}}{4} \\
\Rightarrow {\text{ }}A = {29^ \circ } \\
$
Note: Whenever we face such types of questions, the key concept is that we must covert sin in terms of cos or vice versa. It is clearly visible that here $3A$ represents an acute angle. First we will use the identity $\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)$ then eliminate cos (or sin) then by simplifying the equations we will get our required answer.
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