Question

If the value of $\cos x = \dfrac{1}{2}$ and $\tan B = \dfrac{1}{{\sqrt 3 }}$ , Find $\sin \left( {x + B} \right)$.

Answer 1

Hint: To solve this type of question we must know the concept of trigonometry, its properties and its ratios. Here, firstly we find the value of x as we know that cos ${60^ \circ }$ =$\dfrac{1}{2}$. Similarly, we find the value of B. Thus we find the value of $\sin \left( {x + B} \right)$.

Complete Step-by-Step solution:
Here we are given with $\cos x = \dfrac{1}{2}$
But we know that $\cos {60^ \circ } = \dfrac{1}{2}$
So we get x= ${60^ \circ }$
Also we are given that $\tan B = \dfrac{1}{{\sqrt 3 }}$
But we know that $\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$
So we get B = ${30^ \circ }$
$\sin \left( {x + B} \right) = \sin \left( {{{60}^ \circ } + {{30}^ \circ }} \right)$
$\Rightarrow$ $\sin {90^ \circ }$
And as per trigonometric ratio of sin ${90^ \circ }$ = 1 , so we will get ${90^ \circ }$
Therefore, $\sin \left( {x + B} \right) = 1$

Note: To solve this question we must know all the six trigonometric functions and those are sine function, tangent function, cosecant function, cotangent function, cosine function, secant function. Trigonometry has degrees like ${0^ \circ }, {30^ \circ }, {45^ \circ }, {60^ \circ }, {90^ \circ }$ which has their own varied values which are useful to solve this type of question.