Can someone help me with this geometrical calculus problem concerning trigonometric derivatives?

Question

A semicircle with diameter PQ sits on an isosceles triangle PQR to form a region shaped like a two-dimensional ice cream cone. If A(theta) is the area of the semicircle and B(theta) is the area of the triangle, find: limit, as theta approaches 0 from the right, of A(theta)/B(theta). Sides of the triangle: side PR=10 cm, side QR=10 cm and side PQ forms the bottom of the semicircle. The angle of the bottom of the cone (corner R) is equal to theta.


(this problem is sooo confusing. i will give best answer to the first person to solve it. Thanks!)

Answer 1

PQ = 200 - 200cos(t) = 200[1-cos(t)] = d.

This will make the area of the semicircle = pi*[100(1-cos(t))]^2 = (pi*d^2)/4

The area of the triangle = 1/2*base*height = 1/2*(d/2)*sqrt{100-10000[1-cos(t)]^2} = (1/4)*d*sqrt[100-(d^2/4)].

The ratio of area of semicircle/triangle reduces to (pi*d)/sqrt[100-(d^2/4)].

As t --> 0, d --> 0 so the ratio --> 0. The triangle will have significantly greater area than the semicircle as the angle gets really close to 0, though.

Answer 2

Draw the angle bisector of angle R. It is also the median (and the altitude!) to side PQ. Call the midpoint of PQ as T.

Then PT = QT = radius of the semicircle, so

Area of semicircle = (PT)^2 pi / 2
Area of triangle = (PQ) * (RT) / 2 = (PT) * (RT).

Now, using trig (SOHCAHTOA stuff) you can get

PT / PR = PT / 10 = sin(theta/2)
and
RT / PR = RT / 10 = cos(theta/2).

Solve for PT and RT, then plug those into the area formulas. Then, take the limit. Good luck.

Answer 3

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