Finding area of triangles in circles and taking limits

[sarahjohnson]

Start with a circle of radius r=16. Form the two shaded regions pictured below: c(θ) has an arc and two straight line sides and t(θ) is a right triangle. Note that the areas of these two regions will be functions of θ; r is fixed in the problem.

Find a formula for Area(c(θ))

So my answer is the one above and I believe it to be right because the area of the shaded portion should be the area of the sector which is (theta/360)pi(16^2) and subtract the non shaded triangular portion whose base I used r/2 and the height using the pythagorean theorem. I think the base may be wrong because I don't know for sure if it is 1/2 of r but I do not know how else to find it.

 

[HallsofIvy]

The area of a circular sector with radii r and angle $\displaystyle \theta$ is $\displaystyle \frac{\theta}{2}r^2$. The "base" of the right triangle with angle $\displaystyle \theta$ and hypotenuse r is $\displaystyle r cos(\theta)$ and the height is $\displaystyle r sin(\theta)$ so the area of the triangle is (1/2)(base)(height)= $\displaystyle \frac{1}{2}r^2sin(\theta)cos(\theta)$. This is with the angle \(\displaystyle \theta[/b] in radians. If $\displaystyle \theta$ is in degrees, then the area of the circular sector is $\displaystyle \frac{\theta}{360}\pi r^2$. For the second problem, again the base and height of the "inner triangle" are $\displaystyle rcos(\theta)$ and $\displaystyle rsin(\theta)$, respectively so the height of the triangle you want is the same while its base is $\displaystyle r- rcos(\theta)$. The area of that triangle is $\displaystyle \frac{1}{2}r^2sin(\theta)(1- cos(\theta)$. What limits are to be taken?\)

 

[sarahjohnson]

The area of a circular sector with radii r and angle $\displaystyle \theta$ is $\displaystyle \frac{\theta}{2}r^2$. The "base" of the right triangle with angle $\displaystyle \theta$ and hypotenuse r is $\displaystyle r cos(\theta)$ and the height is $\displaystyle r sin(\theta)$ so the area of the triangle is (1/2)(base)(height)= $\displaystyle \frac{1}{2}r^2sin(\theta)cos(\theta)$. This is with the angle \(\displaystyle \theta[/b] in radians. If $\displaystyle \theta$ is in degrees, then the area of the circular sector is $\displaystyle \frac{\theta}{360}\pi r^2$. For the second problem, again the base and height of the "inner triangle" are $\displaystyle rcos(\theta)$ and $\displaystyle rsin(\theta)$, respectively so the height of the triangle you want is the same while its base is $\displaystyle r- rcos(\theta)$. The area of that triangle is $\displaystyle \frac{1}{2}r^2sin(\theta)(1- cos(\theta)$. What limits are to be taken?\)

\(\displaystyle
Thank you for that explanation. I managed to do the other limits but I do have a problem with the limit as theta goes to zero of Area(c(theta) divided by the area t(theta) which is the first equation divided by the second equation. I thought it should be negative infinity but it says I am wrong.\)