Help please!! T_T
An isosceles triangle is inscribed in a circle of radius R. Determine the value of theta that maximizes the area of the triangle given that theta is the angle contained between the two equal sides.
Help please!! T_T
An isosceles triangle is inscribed in a circle of radius R. Determine the value of theta that maximizes the area of the triangle given that theta is the angle contained between the two equal sides.
I'm not sure what suggesting a height of 2r does for solving the problem, since no triangle inscribed in a circle of radius r could have that height. I would try writing the height and length of the base in terms of trigonometric functions and the radius r.
Ok, first let's establish a name for the triangle. I'll call the circumscribed triangle PQR, where Q has angle theta, and the sides PQ and QR are of the same length. The circle is O. So the lengths of OP, OQ, and OR are all equal to the radius r.
This means that OPQ, OQR, and OPR are all isosceles triangles, with OPQ and OQR congruent. The altitude of triangle PQR divides the angle PQR in half, so angles PQO, RQO, OPQ and ORQ are all equal to one-half theta. This means angles POQ and ROQ are equal to $\displaystyle \pi-\theta$.
With me so far?
Ready or not:
Call the point where the altitude of PQR from Q hits PR point S.
Then angles POS and ROS are both equal to theta. From this, the length of PS is $\displaystyle r sin \theta$ and the length of OS is $\displaystyle r cos \theta$. So the length of altitude QS is $\displaystyle r + r cos \theta$. That's where my area formula comes from. It's the length of PS times the length of QS.