Question

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Answer 1

Given:

The terminal point of $\theta$ is $\left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$.

To find:

The value of $\sin \theta$.

Solution:

If the terminal point of $\theta$ is $(x,y)$, then

$\sin \theta = \dfrac{y}{\sqrt{x^2+y^2}}$

The terminal point of $\theta$ is $\left(\dfrac{1}{2},\dfrac{\sqrt{3}}{2}\right)$. So,

$\sin \theta = \dfrac{\dfrac{\sqrt{3}}{2}}{\sqrt{\left(\dfrac{1}{2}\right)^2+\left(\dfrac{\sqrt{3}}{2}\right)^2}}$

$\sin \theta = \dfrac{\dfrac{\sqrt{3}}{2}}{\sqrt{\dfrac{1}{4}+\dfrac{3}{4}}}$

$\sin \theta = \dfrac{\dfrac{\sqrt{3}}{2}}{\sqrt{1}}$

$\sin \theta = \dfrac{\dfrac{\sqrt{3}}{2}}{1}$

$\sin \theta =\dfrac{\sqrt{3}}{2}$

Therefore, the correct option is B.