Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Based on the figure below, what is the ratio of the area of triangle $CGI$ to the area of triangle $ABC$, in terms of $\theta$?

What is the ratio of the area of triangle $CGI$ to the area of triangle $ABC$, in terms of $\theta$?

share|improve this question
add comment

3 Answers

up vote 6 down vote accepted

All the right triangles in the picture are similar, so (noting $|BG|=|BD|$)

$$\begin{align} \frac{|\triangle CIG|}{|\triangle ABC|} &= \left(\frac{|CG|}{|AC|}\right)^2 = \left(\frac{|BC|-|BG|}{|AC|}\right)^2 = \left(\frac{|BC|-|BD|}{|AC|}\right)^2 \\ &= \left(\frac{|BC|\left(1-\sin\theta\right)}{|AC|}\right)^2 = \left(\frac{|BC|}{|AC|}\right)^2\left(1-\sin\theta\right)^2 \\ &= \tan^2\theta \left(1-\sin\theta\right)^2 \end{align}$$

share|improve this answer
add comment

Fixing $AC$ at length $1$, $\dfrac{\text{Area}(CGI)}{\text{Area}(ABC)}$ is seen to be $\dfrac{\left ( {\dfrac{\sin(\theta)}{\cos(\theta)}} - \sin(\theta)\tan(\theta)\right )^2 \tan(\theta)}{\sin(\theta) \left (\cos(\theta)+ \sin(\theta)\tan(\theta) \right)}$. Mathematica simplifies this ratio to $(\sin(\theta)-1)^2\tan^2(\theta)$.

What is the ratio of the area of triangle $CGI$ to the area of triangle $ABC$, in terms of $\theta$?

share|improve this answer
    
To whomever down-voted this answer: can you explain why you down-voted? –  AbstractionOfMe Jan 27 '13 at 7:36
2  
Looks okay to me. –  Brian M. Scott Jan 27 '13 at 7:44
add comment

Mathematica isn’t needed for the simplification, which is quite straightforward:

$$\begin{align*} \frac{\left({\sin\theta \over \cos\theta} - \sin\theta\tan\theta\right)^2 \tan\theta}{\sin\theta\big(\cos\theta+\sin\theta\tan\theta\big)}&=\frac{\left(\tan\theta-\sin\theta\tan\theta\right)^2\tan\theta}{\frac{\sin\theta}{\cos\theta}\left(\cos^2\theta+\sin^2\theta\right)}\\\\ &=\frac{\tan^3\theta(1-\sin\theta)^2}{\tan\theta}\\\\ &=\tan^2\theta(1-\sin\theta)^2\;. \end{align*}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.