Question

The angle

Answer 1

Answer:

$sin\theta_1 = -\frac{\sqrt{731}}{30}$ is the correct answer.

Step-by-step explanation:

It is given that $\theta_{1}$  is in third quadrant.

$cos\theta_{1}$ is always negative in 3rd quadrant and also

$sin\theta_{1}$ is always negative in 3rd quadrant.

Also, we know the following identity about  and :

$sin^2\theta + cos^2\theta = 1$

Put $\theta$ as $\theta_{1}$:

$sin^2\theta_1 + cos^2\theta_1 = 1$

We are given that $cos\theta_1 = -\frac{13}{30}$

$sin^2\theta_1 + (\frac{-13}{30})^2 = 1$

$\Rightarrow sin^2\theta_1 = 1 - \dfrac{169}{900}\\\Rightarrow sin^2\theta_1 = \dfrac{900-169}{900}\\\Rightarrow sin^2\theta_1 = \dfrac{731}{900}\\\Rightarrow sin\theta_1 = +\sqrt{\dfrac{731}{900}}, -\sqrt{\dfrac{731}{900}}\\\Rightarrow sin\theta_1 = +\dfrac{\sqrt{731}}{30}, -\dfrac{\sqrt{731}}{30}$

$\theta_{1}$ is in 3rd quadrant so $sin\theta_{1}$ is negative.

So $sin\theta_1 = -\frac{\sqrt{731}}{30}$