The problem is as follows:

The figure from below represents a trigonometric circle or unit circle. Assume $PQ=|\sin\theta|$. Find the area of the region represented by $ABC$.

Sketch of the problem

The choices given are as follows:

$\begin{array}{ll} 1.&\frac{\sin\theta}{2}\\ 2.&1\\ 3.&\frac{1}{2}\\ 4.&-\frac{\sin\theta}{2}\\ \end{array}$

Well, I'm totally lost in this problem. As I don't know how to understand the negative value which it has been given in the figure.

The image has been reconstructed from the original source which isn't very clear, so I did to my best of drawing abilities to put all the information but it is not very clear if the angle starts on point which lies on $\textrm{x-axis}$ or in point $B$ so, please consider this in your answer, so it may match the answers given.

In short my biggest complain is that weird arrow how should I understand it?. Can you please address this doubt of mine in the heading of your answer?

Other than that what I could notice is that catch here is how to get the value of the base of the triangle and the height for $\triangle ABC$. The thing is that I'm confused how to use the fact that it confuses me that $PQ$ is given as an absolute value.

Since this was found in an old trigonometry practice sheet the sort of strategy which I'm looking to get is by relying in the concepts of trigonometric circle.

As an extension to this I also notice that the radius here is one unit.

Thus can someone guide me here on exactly what to do, and what should I do to get one of the answers given in the choices?.

Since I'm not very sure with this topic I appreciate that the answer given would include the necessary steps to get to that answer.


The negative value is just a distraction. We only need the magnitude of the angle. $$A_{ABC}=A_{AOC}-A_{AOB} $$ Assuming a unit circle, the base of both triangles, $OA$, is $1$. Also $OQ=1+PQ=1+|sin\theta|$ $$\therefore A_{AOC}=\frac {1\times (1+|sin\theta|)} 2=\frac {1+|sin\theta|} 2 $$ $$A_{AOB}=\frac {1\times altitudeB} 2=\frac {|sin(180-|\theta|)|} 2$$ But $sin(180-\theta)=-sin\theta$ $$\therefore A_{ABC}= \frac {1+|sin\theta|} 2-\frac {|sin\theta|} 2=\frac 1 2$$


Given the configuration, |OA|=1, |OB|=1, $\angle$ GOB=$\theta-90^o$

|OC|=$\frac{|OQ|}{\cos (\theta-90^o)}$=$\frac{(|OP|+|OQ|)}{ \cos (\theta-90^o)}$=$\frac{(1+\sin \theta)}{\sin \theta}$

Apply the area of triangle formula based upon two adjacent sides and the angle in between,

$S_\Delta ABC$ =$S_\Delta AOC$ -$S_\Delta AOB$

= $\frac{1}{2}|OA||OC|\sin(\theta)$-$\frac{1}{2}|OA||OB|\sin(\theta)$

= $\frac{1}{2}\sin\theta$ $\frac{(1+\sin \theta)}{\sin \theta}$-$\frac{1}{2}\sin\theta$

= $\frac{1}{2}$

  • $\begingroup$ Gee First of, thanks for your recent edit. But I'm still confused because the answer doesn't match. Perhaps does it exist a way to match with any?. I mean if a different interpretation of the angle is used?. Let's say if it departs from another position?. About a friend which you mentioned, I don't understand. I own the original source, it is just that it is not very clear so I had to redrawn it and reconstruct it from scratch, so I don't think I can compare it with a clearer version. Have you tried other ways?. $\endgroup$ Mar 13 at 4:00
  • $\begingroup$ @chris Sorry that I made a mistake in |OC| calculation, should have used |OC|=$\frac{|OQ|}{\cos \angle QOC}$, rather than |OQ| cos $\angle QOC$. Everything makes sense now. $\endgroup$ Mar 13 at 9:13

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