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?\)

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