An Isosceles Triangle
Triangle ABC is isosceles, with angle ABC = angle ACB.
The perpendicular bisector of AB intersects BC at point
D, such that DC/BC = 1/3. What is the cosine of angle BAC?
If DC/BC = 1/n, what is the cosine of angle BAC, in terms of n?
Source: Original
Solution:
From Ken Burres:
When DC/BC = 1/3, cos(angle BAC) = 1/4
When DC/BC = 1/n, cos(angle BAC) = (n-2)/(2n-2) in terms of n
Ken Duisenberg's explanation:
Note that triangle BDA is isosceles, with apex D and the same
base angles as the original triangle. So, finding the cosine of
angle BDA will be the same as that of BAC.
Draw the altitude of BAC from
A to E, the midpoint of BC. The cosine of angle BDA is
equal to DE/DA = DE/DB.
Divide BC into 2n divisions. Then CD is 2/2n, so BD is (2n-2)/2n.
DE is (n-2)/2n. So cosine(BDA)=cosine(BAC)=(n-2)/(2n-2).
For n=3, cosine of BAC = 1/4.
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