Circle O has a radius of 3 units. Points A and B are inside the circle such that AO=1 unit, BO=2 unit, and angle AOB=60 degrees.
CDE is an isosceles triangle inscribed in circle O with CD=DE, point A on CD and point B on DE.
How long are the sides of CDE?
This problem is like a circular billiard going from angle to angle to find relations.
Calling a the angle ODB , d the complementary angle DOB (so that the angle DOB is pi-d), and g the angle DBO:
2/sin(a)=3/sin(g)
and g=d-a
from here you get
tan(a)=2sin(d)/(3+2cos(d)) [1]
In addition, calling N the point inside the triangle OAB that makes ANO a right triangle)
tan(a)=AN/ND=AN/(3+NO)=sin(60-d)/(3+cos(60-d) [2]
From [1]=[2], it is derived that:
3sin(60-d)-6sin(d)+2sin(60-2d)=0
Here I prove values, begining from d=25°. Without big precision you get that d=22,7°
Then a=9,051°
CD=(3+3cos(2a))/cos(a)=5.925
CE=2 CD sin(a)=1.864
(the differences with the Sketchpad solutions are because my aproximation to angle d was done only with two decimals).
Edited on March 21, 2019, 10:13 am
|
|
Posted by armando
on 2019-03-21 10:11:38 |
Yes
