I was working on a task given by my Math teacher. The question has to be solved using the following diagram:
The Question is as follows:
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Find the length ${AD}$ in terms of $\beta, \cos\beta,$ and/or $\tan\beta$.
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Find the length ${BC}$ in terms of $\beta, \cos\beta,$ and/or $\tan\beta$.
Another condition is: you cannot use calculus to solve this question (I don't know whether it is required or not though).
My attempt at the question:
While I was trying to solve the question, I was trying to figure out a relation by creating triangles and trying to come up with trigonometric equations using the angles given and the angles which can be figured out (that is, I was trying to see if I can use sin rule, cosine rule, etc.). But all of this was to no avail.
Best Answer
Draw a vertical line from $G$ to the $x$ axis. Call the intersecting point $M$. Find the distance between $O$ and $M$. Length $OM = r \cos\beta$. Now you have the distance from $G$ to the $y$ axis. Draw a vertical line upward from $G$ to create a right triangle with $DG$ as the hypotenuse. Call the intersecting point $N$ between $D$ and $E$. Realize the angle of $GDE$ is $\frac{\pi}{2} - \beta$. The length $DN$ is $\frac{h}{2 \tan(\frac{\pi}{2}-\beta)}$. Subtract the length of $DN$ from the length of $OM$ to get the length of $AD$. Length $AD$ = $r \cos\beta - \frac{h}{2 \tan(\frac{\pi}{2}-\beta)}$.
The length of $BC$ is easily found from here. Draw a vertical line upward from $C$ to intersect with $DE$. The intersection can be called $Z$. The length $DZ$ is twice the length $DN$. Add length $DZ$ to length $AD$ to yield length $BC$. Length $BC = r \cos\beta - \frac{h}{2 \tan(\frac{\pi}{2}-\beta)} + \frac{h}{ \tan(\frac{\pi}{2}-\beta)} = r \cos\beta + \frac{h}{2 \tan(\frac{\pi}{2}-\beta)}$
Edit: Miscellaneous pointed out that $\frac{1}{\tan(\frac{\pi}{2}-\beta)}$ is $\tan\beta$. So, the answer simplifies to: