Given that $\sin heta = \dfrac{3}{5}$ and $ heta$ is obtuse, find $ an heta$.

Question

Given that $\sin	heta = \dfrac{3}{5}$ and $	heta$ is obtuse, find $	an	heta$.

Accepted Answer

$-\dfrac{3}{4}$. From $\sin^2	heta + \cos^2	heta = 1$, $\cos	heta = \pm\dfrac{4}{5}$. Because $	heta$ is obtuse (second quadrant), cosine is negative, so $\cos	heta = -\dfrac{4}{5}$ and $	an	heta = \dfrac{\sin	heta}{\cos	heta} = -\dfrac{3}{4}$. Ignoring the quadrant gives $\dfrac{3}{4}$, and inverting gives

Answer

A) $\dfrac{3}{4}$

Answer

B) $-\dfrac{3}{4}$

Answer

C) $-\dfrac{4}{3}$

Answer

D) $\dfrac{4}{3}$

Given that $\sin\theta = \dfrac{3}{5}$ and $\theta$ is obtuse, find $\tan\theta$.

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