Question

If $\operatorname{Sin}A=\dfrac{1}{3}$ and $A+B=90\text{ degree}$ then find $\cos B$

Answer 1

Hint: We solve this question by using basic trigonometric formulae. We are required to find the value of $\cos B$ and this can be calculated using the relation $\sin \left( 90{}^\circ -\theta \right)=\cos \left( \theta \right).$ This relation is known as the concept of complementary angles. Using this relation along with the given information in the question, we find the value of $\cos B.$

Complete step by step solution:
In order to answer this question, let us note down the given data first. It is given that
$\Rightarrow \sin A=\dfrac{1}{3}\ldots \left( 1 \right)$
It is also given that $A+B=90{}^\circ .$ We need to find the value of $\cos B$ and this can be done by substituting in the equation $\sin A=\dfrac{1}{3}$ in terms of B. So, by representing angle A in terms of B, we can obtain the solution to the above question. Using the second equation,
$\Rightarrow A+B=90{}^\circ $
The above equation represents the concept of complementary angles. Two angles are said to be complementary angles if the sum of the two angles is $90{}^\circ .$ From this equation, we subtract both sides by B.
$\Rightarrow A=90{}^\circ -B$
We now have the angle A represented in terms of B. We substitute this value of A in the equation 1.
$\Rightarrow \sin \left( 90{}^\circ -B \right)=\dfrac{1}{3}$
We use the basic trigonometric relation $\sin \left( 90{}^\circ -\theta \right)=\cos \theta .$ Using this formula for the above equation,
$\Rightarrow \cos B=\dfrac{1}{3}$
Hence, we have obtained the value of $\cos B$ which is given as $\dfrac{1}{3}.$

Note: We need to know the basic trigonometric relations and conversion from sine to cosine functions. It is important to note the conversion from sine to cosine is given by $\sin \left( 90{}^\circ -\theta \right)=\cos \theta $ and vice versa equation is given by $\cos \left( 90{}^\circ -\theta \right)=\sin \theta .$ This concept is called the complementary angles concept. We can use this to solve many mathematical questions.