# The obtuse angle $B$ is such that $\tan B = -\frac{5}{12}$. Find the exact value of $\sin B$.

The obtuse angle $B$ is such that $\tan B = -\frac{5}{12}$. Find the exact value of $\sin B$.

Is this possible to do without a calculator? If so, how?

• @zz20s Wolfram Alpha gives a reasonably nice answer for $\sin(\arctan(-5/12))$. – David K Feb 8 '16 at 14:28
• Without a calculator meaning algebraically...excluding the use of any software like Wolfram Alpha that may not necessarily come under the term calculator... – GoodChessPlayer Feb 8 '16 at 14:33
• Put it into wolfram incorrectly, my fault. – zz20s Feb 8 '16 at 14:35
• @GoodChessPlayer Of course! But the fact that Alpha finds a simple answer suggests that you can find one too without using Alpha. Conversely if Alpha does not find a closed form at all, you probably will not either, since Alpha has all the tricks programmed into it. My actual answer is below. – David K Feb 8 '16 at 14:36

Notice, the obtuse angle $B$ ($90^\circ<B<180^\circ$) lies in second quadrant hence the value of $\sin B$ is positive & is given as follows in terms of $\tan B$,
$$\sin B=\left|\frac{\tan B}{\sqrt{1+\tan^2 B}}\right|=\left|\frac{\frac{-5}{12}}{\sqrt{1+\left(\frac{-5}{12}\right)^2}}\right|=\frac{5}{13}$$

• Nice answer. I think you shall add that you used the identity $\displaystyle \frac{1}{\sin^2(x)}=1+\frac{1}{\tan^2(x)}$. – Galc127 Feb 8 '16 at 14:50

On the Cartesian $x,y$ plane, draw a ray from the origin $(0,0)$ through the point $(-12,5)$. This line makes an angle $\theta$ with the positive $x$-axis (a negative angle, when measured in the counterclockwise direction).

What is $\tan(\theta)$?

How long is the segment from $(0,0)$ to $(-12,5)$?

What is $\sin(\theta)$?

• I do not understand what tan theta would be from such a diagram... – GoodChessPlayer Feb 8 '16 at 14:38
• @GoodChessPlayer What if a triangle has vertices $(0,0)$, $(12,0)$, $(12,5)$? One leg of the triangle has length $12$, the other has length $5$. How can you find the tangent of one angle of a right triangle if you know the length of both legs? (Don't give up until you have at least drawn the diagram!) Now remember that $\tan(\pi-\alpha) = -\tan(\alpha)$. – David K Feb 8 '16 at 16:07